Trapezoidal Sums - AP Calculus BC

Card 0 of 30

Approximate

$\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

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The interval $\left [ 1,2 \right ]$ is 1 unit in width; the interval is divided evenly into five subintervals $\Delta x = 1 \div 5 = 0.2$ units in width. They are

$\left [ 1,1.2 \right ], \left [ 1.2, 1.4 \right ], \left [ 1.4, 1.6 \right ], \left [ 1.6, 1.8 \right ], \left [ 1.8, 2 \right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = 0.2$

$f (x) = e ^${x^{2}$}$

and

.

$T =\frac{0.2}{2}$ \left[ f(1) + 2 f(1.2}) + 2 f(1.4) + 2 f(1.6) + 2 f(1.8) + f(2) \right]$

$f(1) = e ^${1^{2}$} \approx 2.7183$

$f(1.2) = e ^${1.2^{2}$} \approx 4.2207$

$f(1.4) = e ^${1.4^{2}$} \approx 7.0993$

$f(1.6) = e ^${1.6^{2}$} \approx 12.9358$

$f(1.8) = e ^${1.8^{2}$} \approx 25.5337$

$f(2) = e ^${2^{2}$} \approx 54.5982$

So

$T \approx $\frac{0.2}{2}$ \left [2.7183 + 2 \cdot 4.2207 + 2 \cdot 7.0993 + 2 \cdot 12.9358 + 2 \cdot 25.5337 + 54.5982 \right ]$

$\approx 0.1 \left (2.7183 + 8.4414 + 14.1986 + 25.8716 + 51.0674 + 54.5982 \right )$

$\approx 0.1 \cdot 156.8955 \approx 15.690$

Approximate

$\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

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The interval $\left [ 0, $\frac{\pi }{4}\right]$ is $\frac{\pi }{4}$ units in width; the interval is divided evenly into four subintervals $\Delta x = $\frac{\pi }{4}$ \div 4 = $\frac{\pi }{16}$$ units in width. They are

$\left[ 0, $\frac{\pi }{16}$ \right], \left[ $\frac{\pi }{16}$, $\frac{\pi }{8}$ \right], \left[ $\frac{\pi }{8}$, $\frac{3 \pi }{16}$ \right], \left[ $\frac{3 \pi }{16}$, $\frac{\pi }{4}\right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = $\frac{\pi }{16}$$ ,

$f (x) = \tan ^{2}x$ ,

and

.

$T =\frac{\frac{\pi }{16}$}{2} \left[ f(0) + 2 f \left( $\frac{ \pi }{16}\right ) + 2 f \left ( $\frac{ \pi }{8}\right) + 2 f \left( $\frac{ 3 \pi }{16}\right ) + f \left ( $\frac{ \pi }{4}$ \right ) \right ]$

$f(0) = \tan ^{2} 0 = 0$

$f\left ( $\frac{ \pi }{16}\right) = \tan ^{2} $\frac{ \pi }{16}$ \approx 0.0396$

$f\left( $\frac{ \pi }{8}\right ) = \tan ^{2} $\frac{ \pi }{8}$ \approx 0.1716$

$f\left ( $\frac{ 3 \pi }{16}\right) = \tan ^{2} $\frac{ 3 \pi }{16}$ \approx 0.4465$

$f\left( $\frac{ \pi }{4}\right ) = \tan ^{2} $\frac{ \pi }{4}$ = 1$

$$\frac{\frac{\pi }{16}$}{2} = $\frac{\pi }{32}$ = 0.0982$

So

$T \approx 0.0982 \left [0+ 2 \cdot 0.0396 + 2 \cdot 0.1716 + 2 \cdot 0.4465+ 1 \right ]$

$\approx 0.0982 \left [0+ 0.0792 + 0.3432 + 0.8930 + 1 \right ]$

$\approx 0.0982 \cdot 2.3154 \approx 0.227$

Approximate

$\int_${1}^{5}$ $\frac{ \ln x ; \mathrm {d}$x $}{x^2$}$

using the trapezoidal rule with . Round your estimate to three decimal places.

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The interval $\left [ 1,5 \right ]$ is 4 units in width; the interval is divided evenly into four subintervals $\Delta x = 4\div 4 = 1$ units in width - they are .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{5}$ $\frac{ \ln x ; \mathrm {d}$x $}{x^2$}$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + $f(x_{4}$) \right ]$ ,

where $\Delta x = 1$ , $f(x) = $\frac{ \ln x $}{x^2$}$$ , and

.

$T =\frac{1}{2}$ \left[ f(1) + 2 f(2}) + 2 f(3) + 2 f(4) + f(5) \right]$

$f(1) = $\frac{ \ln 1 $}{1^2$}$ = 0$

$f(2) = $\frac{ \ln 2 $}{2^2$}$ \approx $\frac{0.6931 }{4}$ \approx 0.1733$

$f(3) = $\frac{ \ln 3 $}{3^2$}$ \approx $\frac{1.0986 }{9}$ \approx 0.1221$

$f(4) = $\frac{ \ln 4 $}{4^2$}$ \approx $\frac{1.3863 }{16}$ \approx 0.0866$

$f(5) = $\frac{ \ln 5 $}{5^2$}$ \approx $\frac{1.6094 }{25}$ \approx 0.0644$

$T \approx 0.5 \left [0+ 2 \cdot 0.1733 + 2 \cdot 0.1221 + 2 \cdot 0.0866 + 0.0644 \right ]$

$\approx 0.5 \left [ 0.3466 + 0.2442 + 0.1732 + 0.0644 \right ]$

$\approx 0.5 \cdot 0.8284 \approx 0.414$

Approximate

$\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

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The interval $\left [ 1,2 \right ]$ is 1 unit in width; the interval is divided evenly into five subintervals $\Delta x = 1 \div 5 = 0.2$ units in width. They are

$\left [ 1,1.2 \right ], \left [ 1.2, 1.4 \right ], \left [ 1.4, 1.6 \right ], \left [ 1.6, 1.8 \right ], \left [ 1.8, 2 \right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = 0.2$

$f (x) = e ^${x^{2}$}$

and

.

$T =\frac{0.2}{2}$ \left[ f(1) + 2 f(1.2}) + 2 f(1.4) + 2 f(1.6) + 2 f(1.8) + f(2) \right]$

$f(1) = e ^${1^{2}$} \approx 2.7183$

$f(1.2) = e ^${1.2^{2}$} \approx 4.2207$

$f(1.4) = e ^${1.4^{2}$} \approx 7.0993$

$f(1.6) = e ^${1.6^{2}$} \approx 12.9358$

$f(1.8) = e ^${1.8^{2}$} \approx 25.5337$

$f(2) = e ^${2^{2}$} \approx 54.5982$

So

$T \approx $\frac{0.2}{2}$ \left [2.7183 + 2 \cdot 4.2207 + 2 \cdot 7.0993 + 2 \cdot 12.9358 + 2 \cdot 25.5337 + 54.5982 \right ]$

$\approx 0.1 \left (2.7183 + 8.4414 + 14.1986 + 25.8716 + 51.0674 + 54.5982 \right )$

$\approx 0.1 \cdot 156.8955 \approx 15.690$

Approximate

$\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

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The interval $\left [ 0, $\frac{\pi }{4}\right]$ is $\frac{\pi }{4}$ units in width; the interval is divided evenly into four subintervals $\Delta x = $\frac{\pi }{4}$ \div 4 = $\frac{\pi }{16}$$ units in width. They are

$\left[ 0, $\frac{\pi }{16}$ \right], \left[ $\frac{\pi }{16}$, $\frac{\pi }{8}$ \right], \left[ $\frac{\pi }{8}$, $\frac{3 \pi }{16}$ \right], \left[ $\frac{3 \pi }{16}$, $\frac{\pi }{4}\right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = $\frac{\pi }{16}$$ ,

$f (x) = \tan ^{2}x$ ,

and

.

$T =\frac{\frac{\pi }{16}$}{2} \left[ f(0) + 2 f \left( $\frac{ \pi }{16}\right ) + 2 f \left ( $\frac{ \pi }{8}\right) + 2 f \left( $\frac{ 3 \pi }{16}\right ) + f \left ( $\frac{ \pi }{4}$ \right ) \right ]$

$f(0) = \tan ^{2} 0 = 0$

$f\left ( $\frac{ \pi }{16}\right) = \tan ^{2} $\frac{ \pi }{16}$ \approx 0.0396$

$f\left( $\frac{ \pi }{8}\right ) = \tan ^{2} $\frac{ \pi }{8}$ \approx 0.1716$

$f\left ( $\frac{ 3 \pi }{16}\right) = \tan ^{2} $\frac{ 3 \pi }{16}$ \approx 0.4465$

$f\left( $\frac{ \pi }{4}\right ) = \tan ^{2} $\frac{ \pi }{4}$ = 1$

$$\frac{\frac{\pi }{16}$}{2} = $\frac{\pi }{32}$ = 0.0982$

So

$T \approx 0.0982 \left [0+ 2 \cdot 0.0396 + 2 \cdot 0.1716 + 2 \cdot 0.4465+ 1 \right ]$

$\approx 0.0982 \left [0+ 0.0792 + 0.3432 + 0.8930 + 1 \right ]$

$\approx 0.0982 \cdot 2.3154 \approx 0.227$

Approximate

$\int_${1}^{5}$ $\frac{ \ln x ; \mathrm {d}$x $}{x^2$}$

using the trapezoidal rule with . Round your estimate to three decimal places.

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The interval $\left [ 1,5 \right ]$ is 4 units in width; the interval is divided evenly into four subintervals $\Delta x = 4\div 4 = 1$ units in width - they are .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{5}$ $\frac{ \ln x ; \mathrm {d}$x $}{x^2$}$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + $f(x_{4}$) \right ]$ ,

where $\Delta x = 1$ , $f(x) = $\frac{ \ln x $}{x^2$}$$ , and

.

$T =\frac{1}{2}$ \left[ f(1) + 2 f(2}) + 2 f(3) + 2 f(4) + f(5) \right]$

$f(1) = $\frac{ \ln 1 $}{1^2$}$ = 0$

$f(2) = $\frac{ \ln 2 $}{2^2$}$ \approx $\frac{0.6931 }{4}$ \approx 0.1733$

$f(3) = $\frac{ \ln 3 $}{3^2$}$ \approx $\frac{1.0986 }{9}$ \approx 0.1221$

$f(4) = $\frac{ \ln 4 $}{4^2$}$ \approx $\frac{1.3863 }{16}$ \approx 0.0866$

$f(5) = $\frac{ \ln 5 $}{5^2$}$ \approx $\frac{1.6094 }{25}$ \approx 0.0644$

$T \approx 0.5 \left [0+ 2 \cdot 0.1733 + 2 \cdot 0.1221 + 2 \cdot 0.0866 + 0.0644 \right ]$

$\approx 0.5 \left [ 0.3466 + 0.2442 + 0.1732 + 0.0644 \right ]$

$\approx 0.5 \cdot 0.8284 \approx 0.414$

Approximate

$\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

Tap to see back →

The interval $\left [ 1,2 \right ]$ is 1 unit in width; the interval is divided evenly into five subintervals $\Delta x = 1 \div 5 = 0.2$ units in width. They are

$\left [ 1,1.2 \right ], \left [ 1.2, 1.4 \right ], \left [ 1.4, 1.6 \right ], \left [ 1.6, 1.8 \right ], \left [ 1.8, 2 \right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = 0.2$

$f (x) = e ^${x^{2}$}$

and

.

$T =\frac{0.2}{2}$ \left[ f(1) + 2 f(1.2}) + 2 f(1.4) + 2 f(1.6) + 2 f(1.8) + f(2) \right]$

$f(1) = e ^${1^{2}$} \approx 2.7183$

$f(1.2) = e ^${1.2^{2}$} \approx 4.2207$

$f(1.4) = e ^${1.4^{2}$} \approx 7.0993$

$f(1.6) = e ^${1.6^{2}$} \approx 12.9358$

$f(1.8) = e ^${1.8^{2}$} \approx 25.5337$

$f(2) = e ^${2^{2}$} \approx 54.5982$

So

$T \approx $\frac{0.2}{2}$ \left [2.7183 + 2 \cdot 4.2207 + 2 \cdot 7.0993 + 2 \cdot 12.9358 + 2 \cdot 25.5337 + 54.5982 \right ]$

$\approx 0.1 \left (2.7183 + 8.4414 + 14.1986 + 25.8716 + 51.0674 + 54.5982 \right )$

$\approx 0.1 \cdot 156.8955 \approx 15.690$

Approximate

$\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

Tap to see back →

The interval $\left [ 0, $\frac{\pi }{4}\right]$ is $\frac{\pi }{4}$ units in width; the interval is divided evenly into four subintervals $\Delta x = $\frac{\pi }{4}$ \div 4 = $\frac{\pi }{16}$$ units in width. They are

$\left[ 0, $\frac{\pi }{16}$ \right], \left[ $\frac{\pi }{16}$, $\frac{\pi }{8}$ \right], \left[ $\frac{\pi }{8}$, $\frac{3 \pi }{16}$ \right], \left[ $\frac{3 \pi }{16}$, $\frac{\pi }{4}\right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = $\frac{\pi }{16}$$ ,

$f (x) = \tan ^{2}x$ ,

and

.

$T =\frac{\frac{\pi }{16}$}{2} \left[ f(0) + 2 f \left( $\frac{ \pi }{16}\right ) + 2 f \left ( $\frac{ \pi }{8}\right) + 2 f \left( $\frac{ 3 \pi }{16}\right ) + f \left ( $\frac{ \pi }{4}$ \right ) \right ]$

$f(0) = \tan ^{2} 0 = 0$

$f\left ( $\frac{ \pi }{16}\right) = \tan ^{2} $\frac{ \pi }{16}$ \approx 0.0396$

$f\left( $\frac{ \pi }{8}\right ) = \tan ^{2} $\frac{ \pi }{8}$ \approx 0.1716$

$f\left ( $\frac{ 3 \pi }{16}\right) = \tan ^{2} $\frac{ 3 \pi }{16}$ \approx 0.4465$

$f\left( $\frac{ \pi }{4}\right ) = \tan ^{2} $\frac{ \pi }{4}$ = 1$

$$\frac{\frac{\pi }{16}$}{2} = $\frac{\pi }{32}$ = 0.0982$

So

$T \approx 0.0982 \left [0+ 2 \cdot 0.0396 + 2 \cdot 0.1716 + 2 \cdot 0.4465+ 1 \right ]$

$\approx 0.0982 \left [0+ 0.0792 + 0.3432 + 0.8930 + 1 \right ]$

$\approx 0.0982 \cdot 2.3154 \approx 0.227$

Approximate

$\int_${1}^{5}$ $\frac{ \ln x ; \mathrm {d}$x $}{x^2$}$

using the trapezoidal rule with . Round your estimate to three decimal places.

Tap to see back →

The interval $\left [ 1,5 \right ]$ is 4 units in width; the interval is divided evenly into four subintervals $\Delta x = 4\div 4 = 1$ units in width - they are .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{5}$ $\frac{ \ln x ; \mathrm {d}$x $}{x^2$}$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + $f(x_{4}$) \right ]$ ,

where $\Delta x = 1$ , $f(x) = $\frac{ \ln x $}{x^2$}$$ , and

.

$T =\frac{1}{2}$ \left[ f(1) + 2 f(2}) + 2 f(3) + 2 f(4) + f(5) \right]$

$f(1) = $\frac{ \ln 1 $}{1^2$}$ = 0$

$f(2) = $\frac{ \ln 2 $}{2^2$}$ \approx $\frac{0.6931 }{4}$ \approx 0.1733$

$f(3) = $\frac{ \ln 3 $}{3^2$}$ \approx $\frac{1.0986 }{9}$ \approx 0.1221$

$f(4) = $\frac{ \ln 4 $}{4^2$}$ \approx $\frac{1.3863 }{16}$ \approx 0.0866$

$f(5) = $\frac{ \ln 5 $}{5^2$}$ \approx $\frac{1.6094 }{25}$ \approx 0.0644$

$T \approx 0.5 \left [0+ 2 \cdot 0.1733 + 2 \cdot 0.1221 + 2 \cdot 0.0866 + 0.0644 \right ]$

$\approx 0.5 \left [ 0.3466 + 0.2442 + 0.1732 + 0.0644 \right ]$

$\approx 0.5 \cdot 0.8284 \approx 0.414$

Approximate

$\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

Tap to see back →

The interval $\left [ 1,2 \right ]$ is 1 unit in width; the interval is divided evenly into five subintervals $\Delta x = 1 \div 5 = 0.2$ units in width. They are

$\left [ 1,1.2 \right ], \left [ 1.2, 1.4 \right ], \left [ 1.4, 1.6 \right ], \left [ 1.6, 1.8 \right ], \left [ 1.8, 2 \right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = 0.2$

$f (x) = e ^${x^{2}$}$

and

.

$T =\frac{0.2}{2}$ \left[ f(1) + 2 f(1.2}) + 2 f(1.4) + 2 f(1.6) + 2 f(1.8) + f(2) \right]$

$f(1) = e ^${1^{2}$} \approx 2.7183$

$f(1.2) = e ^${1.2^{2}$} \approx 4.2207$

$f(1.4) = e ^${1.4^{2}$} \approx 7.0993$

$f(1.6) = e ^${1.6^{2}$} \approx 12.9358$

$f(1.8) = e ^${1.8^{2}$} \approx 25.5337$

$f(2) = e ^${2^{2}$} \approx 54.5982$

So

$T \approx $\frac{0.2}{2}$ \left [2.7183 + 2 \cdot 4.2207 + 2 \cdot 7.0993 + 2 \cdot 12.9358 + 2 \cdot 25.5337 + 54.5982 \right ]$

$\approx 0.1 \left (2.7183 + 8.4414 + 14.1986 + 25.8716 + 51.0674 + 54.5982 \right )$

$\approx 0.1 \cdot 156.8955 \approx 15.690$

Approximate

$\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

Tap to see back →

The interval $\left [ 0, $\frac{\pi }{4}\right]$ is $\frac{\pi }{4}$ units in width; the interval is divided evenly into four subintervals $\Delta x = $\frac{\pi }{4}$ \div 4 = $\frac{\pi }{16}$$ units in width. They are

$\left[ 0, $\frac{\pi }{16}$ \right], \left[ $\frac{\pi }{16}$, $\frac{\pi }{8}$ \right], \left[ $\frac{\pi }{8}$, $\frac{3 \pi }{16}$ \right], \left[ $\frac{3 \pi }{16}$, $\frac{\pi }{4}\right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = $\frac{\pi }{16}$$ ,

$f (x) = \tan ^{2}x$ ,

and

.

$T =\frac{\frac{\pi }{16}$}{2} \left[ f(0) + 2 f \left( $\frac{ \pi }{16}\right ) + 2 f \left ( $\frac{ \pi }{8}\right) + 2 f \left( $\frac{ 3 \pi }{16}\right ) + f \left ( $\frac{ \pi }{4}$ \right ) \right ]$

$f(0) = \tan ^{2} 0 = 0$

$f\left ( $\frac{ \pi }{16}\right) = \tan ^{2} $\frac{ \pi }{16}$ \approx 0.0396$

$f\left( $\frac{ \pi }{8}\right ) = \tan ^{2} $\frac{ \pi }{8}$ \approx 0.1716$

$f\left ( $\frac{ 3 \pi }{16}\right) = \tan ^{2} $\frac{ 3 \pi }{16}$ \approx 0.4465$

$f\left( $\frac{ \pi }{4}\right ) = \tan ^{2} $\frac{ \pi }{4}$ = 1$

$$\frac{\frac{\pi }{16}$}{2} = $\frac{\pi }{32}$ = 0.0982$

So

$T \approx 0.0982 \left [0+ 2 \cdot 0.0396 + 2 \cdot 0.1716 + 2 \cdot 0.4465+ 1 \right ]$

$\approx 0.0982 \left [0+ 0.0792 + 0.3432 + 0.8930 + 1 \right ]$

$\approx 0.0982 \cdot 2.3154 \approx 0.227$

Approximate

$\int_${1}^{5}$ $\frac{ \ln x ; \mathrm {d}$x $}{x^2$}$

using the trapezoidal rule with . Round your estimate to three decimal places.

Tap to see back →

The interval $\left [ 1,5 \right ]$ is 4 units in width; the interval is divided evenly into four subintervals $\Delta x = 4\div 4 = 1$ units in width - they are .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{5}$ $\frac{ \ln x ; \mathrm {d}$x $}{x^2$}$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + $f(x_{4}$) \right ]$ ,

where $\Delta x = 1$ , $f(x) = $\frac{ \ln x $}{x^2$}$$ , and

.

$T =\frac{1}{2}$ \left[ f(1) + 2 f(2}) + 2 f(3) + 2 f(4) + f(5) \right]$

$f(1) = $\frac{ \ln 1 $}{1^2$}$ = 0$

$f(2) = $\frac{ \ln 2 $}{2^2$}$ \approx $\frac{0.6931 }{4}$ \approx 0.1733$

$f(3) = $\frac{ \ln 3 $}{3^2$}$ \approx $\frac{1.0986 }{9}$ \approx 0.1221$

$f(4) = $\frac{ \ln 4 $}{4^2$}$ \approx $\frac{1.3863 }{16}$ \approx 0.0866$

$f(5) = $\frac{ \ln 5 $}{5^2$}$ \approx $\frac{1.6094 }{25}$ \approx 0.0644$

$T \approx 0.5 \left [0+ 2 \cdot 0.1733 + 2 \cdot 0.1221 + 2 \cdot 0.0866 + 0.0644 \right ]$

$\approx 0.5 \left [ 0.3466 + 0.2442 + 0.1732 + 0.0644 \right ]$

$\approx 0.5 \cdot 0.8284 \approx 0.414$

Approximate

$\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

Tap to see back →

The interval $\left [ 1,2 \right ]$ is 1 unit in width; the interval is divided evenly into five subintervals $\Delta x = 1 \div 5 = 0.2$ units in width. They are

$\left [ 1,1.2 \right ], \left [ 1.2, 1.4 \right ], \left [ 1.4, 1.6 \right ], \left [ 1.6, 1.8 \right ], \left [ 1.8, 2 \right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = 0.2$

$f (x) = e ^${x^{2}$}$

and

.

$T =\frac{0.2}{2}$ \left[ f(1) + 2 f(1.2}) + 2 f(1.4) + 2 f(1.6) + 2 f(1.8) + f(2) \right]$

$f(1) = e ^${1^{2}$} \approx 2.7183$

$f(1.2) = e ^${1.2^{2}$} \approx 4.2207$

$f(1.4) = e ^${1.4^{2}$} \approx 7.0993$

$f(1.6) = e ^${1.6^{2}$} \approx 12.9358$

$f(1.8) = e ^${1.8^{2}$} \approx 25.5337$

$f(2) = e ^${2^{2}$} \approx 54.5982$

So

$T \approx $\frac{0.2}{2}$ \left [2.7183 + 2 \cdot 4.2207 + 2 \cdot 7.0993 + 2 \cdot 12.9358 + 2 \cdot 25.5337 + 54.5982 \right ]$

$\approx 0.1 \left (2.7183 + 8.4414 + 14.1986 + 25.8716 + 51.0674 + 54.5982 \right )$

$\approx 0.1 \cdot 156.8955 \approx 15.690$

Approximate

$\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

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The interval $\left [ 0, $\frac{\pi }{4}\right]$ is $\frac{\pi }{4}$ units in width; the interval is divided evenly into four subintervals $\Delta x = $\frac{\pi }{4}$ \div 4 = $\frac{\pi }{16}$$ units in width. They are

$\left[ 0, $\frac{\pi }{16}$ \right], \left[ $\frac{\pi }{16}$, $\frac{\pi }{8}$ \right], \left[ $\frac{\pi }{8}$, $\frac{3 \pi }{16}$ \right], \left[ $\frac{3 \pi }{16}$, $\frac{\pi }{4}\right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = $\frac{\pi }{16}$$ ,

$f (x) = \tan ^{2}x$ ,

and

.

$T =\frac{\frac{\pi }{16}$}{2} \left[ f(0) + 2 f \left( $\frac{ \pi }{16}\right ) + 2 f \left ( $\frac{ \pi }{8}\right) + 2 f \left( $\frac{ 3 \pi }{16}\right ) + f \left ( $\frac{ \pi }{4}$ \right ) \right ]$

$f(0) = \tan ^{2} 0 = 0$

$f\left ( $\frac{ \pi }{16}\right) = \tan ^{2} $\frac{ \pi }{16}$ \approx 0.0396$

$f\left( $\frac{ \pi }{8}\right ) = \tan ^{2} $\frac{ \pi }{8}$ \approx 0.1716$

$f\left ( $\frac{ 3 \pi }{16}\right) = \tan ^{2} $\frac{ 3 \pi }{16}$ \approx 0.4465$

$f\left( $\frac{ \pi }{4}\right ) = \tan ^{2} $\frac{ \pi }{4}$ = 1$

$$\frac{\frac{\pi }{16}$}{2} = $\frac{\pi }{32}$ = 0.0982$

So

$T \approx 0.0982 \left [0+ 2 \cdot 0.0396 + 2 \cdot 0.1716 + 2 \cdot 0.4465+ 1 \right ]$

$\approx 0.0982 \left [0+ 0.0792 + 0.3432 + 0.8930 + 1 \right ]$

$\approx 0.0982 \cdot 2.3154 \approx 0.227$

Approximate

$\int_${1}^{5}$ $\frac{ \ln x ; \mathrm {d}$x $}{x^2$}$

using the trapezoidal rule with . Round your estimate to three decimal places.

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The interval $\left [ 1,5 \right ]$ is 4 units in width; the interval is divided evenly into four subintervals $\Delta x = 4\div 4 = 1$ units in width - they are .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{5}$ $\frac{ \ln x ; \mathrm {d}$x $}{x^2$}$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + $f(x_{4}$) \right ]$ ,

where $\Delta x = 1$ , $f(x) = $\frac{ \ln x $}{x^2$}$$ , and

.

$T =\frac{1}{2}$ \left[ f(1) + 2 f(2}) + 2 f(3) + 2 f(4) + f(5) \right]$

$f(1) = $\frac{ \ln 1 $}{1^2$}$ = 0$

$f(2) = $\frac{ \ln 2 $}{2^2$}$ \approx $\frac{0.6931 }{4}$ \approx 0.1733$

$f(3) = $\frac{ \ln 3 $}{3^2$}$ \approx $\frac{1.0986 }{9}$ \approx 0.1221$

$f(4) = $\frac{ \ln 4 $}{4^2$}$ \approx $\frac{1.3863 }{16}$ \approx 0.0866$

$f(5) = $\frac{ \ln 5 $}{5^2$}$ \approx $\frac{1.6094 }{25}$ \approx 0.0644$

$T \approx 0.5 \left [0+ 2 \cdot 0.1733 + 2 \cdot 0.1221 + 2 \cdot 0.0866 + 0.0644 \right ]$

$\approx 0.5 \left [ 0.3466 + 0.2442 + 0.1732 + 0.0644 \right ]$

$\approx 0.5 \cdot 0.8284 \approx 0.414$

Approximate

$\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

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The interval $\left [ 1,2 \right ]$ is 1 unit in width; the interval is divided evenly into five subintervals $\Delta x = 1 \div 5 = 0.2$ units in width. They are

$\left [ 1,1.2 \right ], \left [ 1.2, 1.4 \right ], \left [ 1.4, 1.6 \right ], \left [ 1.6, 1.8 \right ], \left [ 1.8, 2 \right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = 0.2$

$f (x) = e ^${x^{2}$}$

and

.

$T =\frac{0.2}{2}$ \left[ f(1) + 2 f(1.2}) + 2 f(1.4) + 2 f(1.6) + 2 f(1.8) + f(2) \right]$

$f(1) = e ^${1^{2}$} \approx 2.7183$

$f(1.2) = e ^${1.2^{2}$} \approx 4.2207$

$f(1.4) = e ^${1.4^{2}$} \approx 7.0993$

$f(1.6) = e ^${1.6^{2}$} \approx 12.9358$

$f(1.8) = e ^${1.8^{2}$} \approx 25.5337$

$f(2) = e ^${2^{2}$} \approx 54.5982$

So

$T \approx $\frac{0.2}{2}$ \left [2.7183 + 2 \cdot 4.2207 + 2 \cdot 7.0993 + 2 \cdot 12.9358 + 2 \cdot 25.5337 + 54.5982 \right ]$

$\approx 0.1 \left (2.7183 + 8.4414 + 14.1986 + 25.8716 + 51.0674 + 54.5982 \right )$

$\approx 0.1 \cdot 156.8955 \approx 15.690$

Approximate

$\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

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The interval $\left [ 0, $\frac{\pi }{4}\right]$ is $\frac{\pi }{4}$ units in width; the interval is divided evenly into four subintervals $\Delta x = $\frac{\pi }{4}$ \div 4 = $\frac{\pi }{16}$$ units in width. They are

$\left[ 0, $\frac{\pi }{16}$ \right], \left[ $\frac{\pi }{16}$, $\frac{\pi }{8}$ \right], \left[ $\frac{\pi }{8}$, $\frac{3 \pi }{16}$ \right], \left[ $\frac{3 \pi }{16}$, $\frac{\pi }{4}\right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = $\frac{\pi }{16}$$ ,

$f (x) = \tan ^{2}x$ ,

and

.

$T =\frac{\frac{\pi }{16}$}{2} \left[ f(0) + 2 f \left( $\frac{ \pi }{16}\right ) + 2 f \left ( $\frac{ \pi }{8}\right) + 2 f \left( $\frac{ 3 \pi }{16}\right ) + f \left ( $\frac{ \pi }{4}$ \right ) \right ]$

$f(0) = \tan ^{2} 0 = 0$

$f\left ( $\frac{ \pi }{16}\right) = \tan ^{2} $\frac{ \pi }{16}$ \approx 0.0396$

$f\left( $\frac{ \pi }{8}\right ) = \tan ^{2} $\frac{ \pi }{8}$ \approx 0.1716$

$f\left ( $\frac{ 3 \pi }{16}\right) = \tan ^{2} $\frac{ 3 \pi }{16}$ \approx 0.4465$

$f\left( $\frac{ \pi }{4}\right ) = \tan ^{2} $\frac{ \pi }{4}$ = 1$

$$\frac{\frac{\pi }{16}$}{2} = $\frac{\pi }{32}$ = 0.0982$

So

$T \approx 0.0982 \left [0+ 2 \cdot 0.0396 + 2 \cdot 0.1716 + 2 \cdot 0.4465+ 1 \right ]$

$\approx 0.0982 \left [0+ 0.0792 + 0.3432 + 0.8930 + 1 \right ]$

$\approx 0.0982 \cdot 2.3154 \approx 0.227$

Approximate

$\int_${1}^{5}$ $\frac{ \ln x ; \mathrm {d}$x $}{x^2$}$

using the trapezoidal rule with . Round your estimate to three decimal places.

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The interval $\left [ 1,5 \right ]$ is 4 units in width; the interval is divided evenly into four subintervals $\Delta x = 4\div 4 = 1$ units in width - they are .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{5}$ $\frac{ \ln x ; \mathrm {d}$x $}{x^2$}$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + $f(x_{4}$) \right ]$ ,

where $\Delta x = 1$ , $f(x) = $\frac{ \ln x $}{x^2$}$$ , and

.

$T =\frac{1}{2}$ \left[ f(1) + 2 f(2}) + 2 f(3) + 2 f(4) + f(5) \right]$

$f(1) = $\frac{ \ln 1 $}{1^2$}$ = 0$

$f(2) = $\frac{ \ln 2 $}{2^2$}$ \approx $\frac{0.6931 }{4}$ \approx 0.1733$

$f(3) = $\frac{ \ln 3 $}{3^2$}$ \approx $\frac{1.0986 }{9}$ \approx 0.1221$

$f(4) = $\frac{ \ln 4 $}{4^2$}$ \approx $\frac{1.3863 }{16}$ \approx 0.0866$

$f(5) = $\frac{ \ln 5 $}{5^2$}$ \approx $\frac{1.6094 }{25}$ \approx 0.0644$

$T \approx 0.5 \left [0+ 2 \cdot 0.1733 + 2 \cdot 0.1221 + 2 \cdot 0.0866 + 0.0644 \right ]$

$\approx 0.5 \left [ 0.3466 + 0.2442 + 0.1732 + 0.0644 \right ]$

$\approx 0.5 \cdot 0.8284 \approx 0.414$

Approximate

$\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

Tap to see back →

The interval $\left [ 1,2 \right ]$ is 1 unit in width; the interval is divided evenly into five subintervals $\Delta x = 1 \div 5 = 0.2$ units in width. They are

$\left [ 1,1.2 \right ], \left [ 1.2, 1.4 \right ], \left [ 1.4, 1.6 \right ], \left [ 1.6, 1.8 \right ], \left [ 1.8, 2 \right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${1}^{2}$ e ^${x^{2}$} \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = 0.2$

$f (x) = e ^${x^{2}$}$

and

.

$T =\frac{0.2}{2}$ \left[ f(1) + 2 f(1.2}) + 2 f(1.4) + 2 f(1.6) + 2 f(1.8) + f(2) \right]$

$f(1) = e ^${1^{2}$} \approx 2.7183$

$f(1.2) = e ^${1.2^{2}$} \approx 4.2207$

$f(1.4) = e ^${1.4^{2}$} \approx 7.0993$

$f(1.6) = e ^${1.6^{2}$} \approx 12.9358$

$f(1.8) = e ^${1.8^{2}$} \approx 25.5337$

$f(2) = e ^${2^{2}$} \approx 54.5982$

So

$T \approx $\frac{0.2}{2}$ \left [2.7183 + 2 \cdot 4.2207 + 2 \cdot 7.0993 + 2 \cdot 12.9358 + 2 \cdot 25.5337 + 54.5982 \right ]$

$\approx 0.1 \left (2.7183 + 8.4414 + 14.1986 + 25.8716 + 51.0674 + 54.5982 \right )$

$\approx 0.1 \cdot 156.8955 \approx 15.690$

Approximate

$\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$

using the trapezoidal rule with . Round your answer to three decimal places.

Tap to see back →

The interval $\left [ 0, $\frac{\pi }{4}\right]$ is $\frac{\pi }{4}$ units in width; the interval is divided evenly into four subintervals $\Delta x = $\frac{\pi }{4}$ \div 4 = $\frac{\pi }{16}$$ units in width. They are

$\left[ 0, $\frac{\pi }{16}$ \right], \left[ $\frac{\pi }{16}$, $\frac{\pi }{8}$ \right], \left[ $\frac{\pi }{8}$, $\frac{3 \pi }{16}$ \right], \left[ $\frac{3 \pi }{16}$, $\frac{\pi }{4}\right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_${0}^{ $\frac{\pi}${4}$} \tan ^{2}x ; \mathrm{d}x$ by evaluating

$T = $\frac{\Delta x}{2}$ \left [ $f(x_{0}$) + 2 $f(x_{1}$) + 2 $f(x_{2}$) + 2 $f(x_{3}$) + 2 $f(x_{4}$) + $f(x_{5}$) \right ]$ ,

where

$\Delta x = $\frac{\pi }{16}$$ ,

$f (x) = \tan ^{2}x$ ,

and

.

$T =\frac{\frac{\pi }{16}$}{2} \left[ f(0) + 2 f \left( $\frac{ \pi }{16}\right ) + 2 f \left ( $\frac{ \pi }{8}\right) + 2 f \left( $\frac{ 3 \pi }{16}\right ) + f \left ( $\frac{ \pi }{4}$ \right ) \right ]$

$f(0) = \tan ^{2} 0 = 0$

$f\left ( $\frac{ \pi }{16}\right) = \tan ^{2} $\frac{ \pi }{16}$ \approx 0.0396$

$f\left( $\frac{ \pi }{8}\right ) = \tan ^{2} $\frac{ \pi }{8}$ \approx 0.1716$

$f\left ( $\frac{ 3 \pi }{16}\right) = \tan ^{2} $\frac{ 3 \pi }{16}$ \approx 0.4465$

$f\left( $\frac{ \pi }{4}\right ) = \tan ^{2} $\frac{ \pi }{4}$ = 1$

$$\frac{\frac{\pi }{16}$}{2} = $\frac{\pi }{32}$ = 0.0982$

So

$T \approx 0.0982 \left [0+ 2 \cdot 0.0396 + 2 \cdot 0.1716 + 2 \cdot 0.4465+ 1 \right ]$

$\approx 0.0982 \left [0+ 0.0792 + 0.3432 + 0.8930 + 1 \right ]$

$\approx 0.0982 \cdot 2.3154 \approx 0.227$