Fundamental Theorem of Calculus with Definite Integrals - AP Calculus BC

Card 0 of 110

Evaluate :

$f(x)=\int_${0}^{x}$ $\sin\left($\sqrt{t^2$$+\frac{\pi^2$}{4}$} \right )dt$

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By the Fundamental Theorem of Calculus, we have that ${f'(x) = $\frac{d}{dx}$ \int_$0^x$ \sin $$\sqrt{t^2$ + \frac{\pi ^2}{4}$} dt = \sin $$\sqrt{x^2$ + \frac{\pi ^2}{4}$}$ . Thus, $f'(0) = \sin $$\sqrt{0^2$ + \frac{\pi ^2}{4}$} =1$ .

Find the result:

$$\frac{\mathrm{d}$ }{\mathrm{d} x} \int_${1}^{e^x}$ t \ln t ; \mathrm{d}t$

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Set . Then $\frac{\mathrm{d}$ u}{\mathrm{d} x} = $e^x$ , and by the chain rule,

$$\frac{\mathrm{d}$ }{\mathrm{d} x} \int_${1}^{e^x}$ t \ln t ; \mathrm{d}t$

$=\frac{\mathrm{d}$ u}{\mathrm{d} x} \cdot $\frac{\mathrm{d}$ }{\mathrm{d} u} \int_${1}^{u}$ t \ln t ; \mathrm{d}t$

$= $e^x$ \cdot $\frac{\mathrm{d}$ }{\mathrm{d} u} \int_${1}^{u}$ t \ln t ; \mathrm{d}t$

By the fundamental theorem of Calculus, the above can be rewritten as

$= $e^x$ \cdot $e^x$ \ln $e^x$$

$= $e^{2x}$ \cdot x = $xe^{2x}$$

Suppose we have the function

$\small g(x)=\int_${0}^{\cos x}$(1+\sin t)dt$

What is the derivative, $\small g'(x)$ ?

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We can view the function $\small g(x)$ as a function of $\small \cos x$ , as so

$\small g(x)=F(\cos x)$

where $\small F(x)=\int_$0^x$ (1+\sin t)dt$ .

We can find the derivative of $\small g(x)$ using the chain rule:

$\small g'(x)=F'(\cos x)\cdot (\cos x)'=F'(\cos x)\cdot (-\sin x)$

where $\small F'(z)$ can be found using the fundamental theorem of calculus:

$\small \small F'(x)=\frac{d}{dz}$\int_$0^z$ (1+\sin t)dt=1+\sin z$

So we get

$\small \small \small \small g'(x)=-\sin x\cdot F'(\cos x)=-\sin x\cdot[1+\sin(\cos x)]$

Evaluate when $f(x)=\int_${0}^{x}$$(2t^{2}$+5t-4)dt$ .

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Via the Fundamental Theorem of Calculus, we know that, given a function $f(x)=\int_${0}^{x}$$(2t^{2}$+5t-4)dt$ , $f'(x)=\frac{d}{dx}$\int_${0}^{x}$$(2t^{2}$$+5t-4)dt=2x^{2}$+5x-4$ .

Therefore .

Evaluate when $f(x)=\int_${0}^{x}$$(6t^{2}$-3t+10)dt$ .

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Via the Fundamental Theorem of Calculus, we know that, given a function $f(x)=\int_${0}^{x}$$(6t^{2}$-3t+10)dt$ , $f'(x)=\frac{d}{dx}$\int_${0}^{x}$$(6t^{2}$$-3t+10)dt=6x^{2}$-3x+10$ . Therefore, .

Given

$f(x)=\int_${0}^{x}$$\left($\frac{1}{t^{2}$$}-2t+3\right)dt$ , what is ?

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By the Fundamental Theorem of Calculus, for all functions that are continuously defined on the interval with in and for all functions defined by by $F(x)=\int_${a}^{x}$f(t)dt$ , we know that .

Thus, for

$f(x)=\int_${0}^{x}$$\left($\frac{1}{t^{2}$$}-2t+3\right)dt$ ,

$f'(x)=\frac{d}{dx}$\int_${0}^{x}$$\left($\frac{1}{t^{2}$$$}-2t+3\right)dt=\frac{1}{x^{2}$$}-2x+3$ .

Therefore,

Given

$f(x)=\int_${0}^{x}$$($\frac{2}{t^{2}$$+4t-5})dt$ , what is ?

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By the Fundamental Theorem of Calculus, for all functions that are continuously defined on the interval with in and for all functions defined by by $F(x)=\int_${a}^{x}$f(t)dt$ , we know that .

Given

$f(x)=\int_${0}^{x}$$\left($\frac{2}{t^{2}$$+4t-5}\right)dt$ , then

$f'(x)=\frac{d}{dx}$\int_${0}^{x}$$\left($\frac{2}{t^{2}$$$+4t-5}\right)dt=\frac{2}{x^{2}$$+4x-5}$ .

Therefore,

.

Evaluate $\int_${0}^{6}$ $(x-3)^2$ dx$

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$\int $(x-3)^2$ dx = $\frac{1}{3}$ $(x-3)^3$ + C$

Use the fundamental theorem of calculus to evaluate:

$\frac{1}{3}$ $(6-3)^3$ - [ $$\frac{1}{3}$(0-3)^3$] = $\frac{1}{3}$ $(3)^3$ - $\frac{1}{3}$ $(-3)^3$

$9 + $\frac{1}{3}$(27) = 9+9 = 18$

$\int _{$\frac{\pi}{3}$} ^ {\pi } \sin (x/2) dx$

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$\int \sin (x/2) dx = -2 \cos (x/2 )$

Use the Fundamental Theorem of Calculus and evaluate the integral at both endpoints:

$-2 \cos (\pi/2 ) - -2 \cos ($\frac{\pi}{3}$ / 2 ) = -2 \cos ($\frac{\pi}{2}$) + 2 \cos ($\frac{\pi}{6}$)$

$= -2(0) + 2($\frac{\sqrt3}{2}$ )= \sqrt3$

$\int_${-1}^{3}$ \left $(4x^2$ - x \right )dx$

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$\int \left ( $4x^2$ -x\right ) dx = $\frac{4}{3}$ $x^3$ - $\frac{1}{2}$ x + C$

Use the Fundamental Theorem of Calculus and evaluate the integral at both endpoints:

$$\frac{4}{3}$ $(3)^3$ - $\frac{1}{2}$ $(3)^2$ - [ $\frac{4}{3}$ $(-1)^3$ - $\frac{1}{2}$ $(-1)^2$]$

$= 36 - 4.5 - [ -$\frac{4}{3}$ - $\frac{1}{2}$ ] = 36 - 4.5 + 1. $\overline{3}$ + .5 = 33. $\overline{3}$$

Evaluate the following integral

$\int_$0^5$ $3x^4$$-12x^3$+10x+12dx$

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Evaluate the following integral

$\int_$0^5$ $3x^4$$-12x^3$+10x+12dx$

Let's begin by recalling our "reverse power rule" AKA, the antiderivative form of our power rule.

$\int $x^ndx$$=\frac{x^{n+1}$$}{n+1}+c$

In other words, all we need to do for each term is increase the exponent by 1 and then divide by that number.

$\int_$0^5$ $3x^4$$-12x^3$$+10x+12dx=\frac{3x^5$$}{5}$-$\frac{12x^4$$}{4}$+\frac{10x^2$}{2}$+12x+c\mid_$0^5$$

Let's clean it up a little to get:

Now, to evaluate our integral, we need to plug in 5 and 0 for x and find the difference between the values. In other words, if our integrated function is F(x), we need to find F(5)-F(0).

Let's start with F(5)

Next, let's look at F(0). If you look at our function carefully, you will notice that F(0) will cancel out all of our terms except for +c. So, we have the following:

Finding the difference cancels out the c's and leaves us with 185.

Evaluate :

$f(x)=\int_${0}^{x}$ $\sin\left($\sqrt{t^2$$+\frac{\pi^2$}{4}$} \right )dt$

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By the Fundamental Theorem of Calculus, we have that ${f'(x) = $\frac{d}{dx}$ \int_$0^x$ \sin $$\sqrt{t^2$ + \frac{\pi ^2}{4}$} dt = \sin $$\sqrt{x^2$ + \frac{\pi ^2}{4}$}$ . Thus, $f'(0) = \sin $$\sqrt{0^2$ + \frac{\pi ^2}{4}$} =1$ .

Find the result:

$$\frac{\mathrm{d}$ }{\mathrm{d} x} \int_${1}^{e^x}$ t \ln t ; \mathrm{d}t$

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Set . Then $\frac{\mathrm{d}$ u}{\mathrm{d} x} = $e^x$ , and by the chain rule,

$$\frac{\mathrm{d}$ }{\mathrm{d} x} \int_${1}^{e^x}$ t \ln t ; \mathrm{d}t$

$=\frac{\mathrm{d}$ u}{\mathrm{d} x} \cdot $\frac{\mathrm{d}$ }{\mathrm{d} u} \int_${1}^{u}$ t \ln t ; \mathrm{d}t$

$= $e^x$ \cdot $\frac{\mathrm{d}$ }{\mathrm{d} u} \int_${1}^{u}$ t \ln t ; \mathrm{d}t$

By the fundamental theorem of Calculus, the above can be rewritten as

$= $e^x$ \cdot $e^x$ \ln $e^x$$

$= $e^{2x}$ \cdot x = $xe^{2x}$$

Suppose we have the function

$\small g(x)=\int_${0}^{\cos x}$(1+\sin t)dt$

What is the derivative, $\small g'(x)$ ?

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We can view the function $\small g(x)$ as a function of $\small \cos x$ , as so

$\small g(x)=F(\cos x)$

where $\small F(x)=\int_$0^x$ (1+\sin t)dt$ .

We can find the derivative of $\small g(x)$ using the chain rule:

$\small g'(x)=F'(\cos x)\cdot (\cos x)'=F'(\cos x)\cdot (-\sin x)$

where $\small F'(z)$ can be found using the fundamental theorem of calculus:

$\small \small F'(x)=\frac{d}{dz}$\int_$0^z$ (1+\sin t)dt=1+\sin z$

So we get

$\small \small \small \small g'(x)=-\sin x\cdot F'(\cos x)=-\sin x\cdot[1+\sin(\cos x)]$

Evaluate when $f(x)=\int_${0}^{x}$$(2t^{2}$+5t-4)dt$ .

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Via the Fundamental Theorem of Calculus, we know that, given a function $f(x)=\int_${0}^{x}$$(2t^{2}$+5t-4)dt$ , $f'(x)=\frac{d}{dx}$\int_${0}^{x}$$(2t^{2}$$+5t-4)dt=2x^{2}$+5x-4$ .

Therefore .

Evaluate when $f(x)=\int_${0}^{x}$$(6t^{2}$-3t+10)dt$ .

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Via the Fundamental Theorem of Calculus, we know that, given a function $f(x)=\int_${0}^{x}$$(6t^{2}$-3t+10)dt$ , $f'(x)=\frac{d}{dx}$\int_${0}^{x}$$(6t^{2}$$-3t+10)dt=6x^{2}$-3x+10$ . Therefore, .

Given

$f(x)=\int_${0}^{x}$$\left($\frac{1}{t^{2}$$}-2t+3\right)dt$ , what is ?

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By the Fundamental Theorem of Calculus, for all functions that are continuously defined on the interval with in and for all functions defined by by $F(x)=\int_${a}^{x}$f(t)dt$ , we know that .

Thus, for

$f(x)=\int_${0}^{x}$$\left($\frac{1}{t^{2}$$}-2t+3\right)dt$ ,

$f'(x)=\frac{d}{dx}$\int_${0}^{x}$$\left($\frac{1}{t^{2}$$$}-2t+3\right)dt=\frac{1}{x^{2}$$}-2x+3$ .

Therefore,

Given

$f(x)=\int_${0}^{x}$$($\frac{2}{t^{2}$$+4t-5})dt$ , what is ?

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By the Fundamental Theorem of Calculus, for all functions that are continuously defined on the interval with in and for all functions defined by by $F(x)=\int_${a}^{x}$f(t)dt$ , we know that .

Given

$f(x)=\int_${0}^{x}$$\left($\frac{2}{t^{2}$$+4t-5}\right)dt$ , then

$f'(x)=\frac{d}{dx}$\int_${0}^{x}$$\left($\frac{2}{t^{2}$$$+4t-5}\right)dt=\frac{2}{x^{2}$$+4x-5}$ .

Therefore,

.

Evaluate $\int_${0}^{6}$ $(x-3)^2$ dx$

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$\int $(x-3)^2$ dx = $\frac{1}{3}$ $(x-3)^3$ + C$

Use the fundamental theorem of calculus to evaluate:

$\frac{1}{3}$ $(6-3)^3$ - [ $$\frac{1}{3}$(0-3)^3$] = $\frac{1}{3}$ $(3)^3$ - $\frac{1}{3}$ $(-3)^3$

$9 + $\frac{1}{3}$(27) = 9+9 = 18$

$\int _{$\frac{\pi}{3}$} ^ {\pi } \sin (x/2) dx$

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$\int \sin (x/2) dx = -2 \cos (x/2 )$

Use the Fundamental Theorem of Calculus and evaluate the integral at both endpoints:

$-2 \cos (\pi/2 ) - -2 \cos ($\frac{\pi}{3}$ / 2 ) = -2 \cos ($\frac{\pi}{2}$) + 2 \cos ($\frac{\pi}{6}$)$

$= -2(0) + 2($\frac{\sqrt3}{2}$ )= \sqrt3$