Fundamental Theorem of Calculus and Techniques of Antidifferentiation - AP Calculus BC

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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:

$\int_${0}^{\infty }$ $\frac{\cos x; $\mathrm{d}$x}{e^{x}$}$

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First, we will find the indefinite integral $\int $\frac{\cos x; $\mathrm{d}$x}{e^{x}$} = \int $e^{-x}$ \cos x; \mathrm{d}x$ using integration by parts.

We will let $u = \cos x$ and $\mathrm{d}v= $e^{-x}$; \mathrm{d}x$ .

Then $$\frac{\mathrm{d}$ u}{\mathrm{d} x} = - \sin x$ and $\mathrm{d} u = - \sin x ; \mathrm{d} x$ .

$v= \int $e^{-x}$; \mathrm{d}x = - $e^{-x}$ = - $$\frac{1}{e^{x}$$ }$

$\int $\frac{\cos x; $\mathrm{d}$x}{e^{x}$} = \int $e^{-x}$ \cos x; \mathrm{d}x$

$=\int u ; \mathrm{d} v$

$= uv - \int v ; \mathrm{d} u$

$= \cos x \cdot \left (- $$\frac{1}{e^{x}$$ } \right ) - \int \left (- $$\frac{1}{e^{x}$$ } \right ) ; \cdot \left (- \sin x \right ); \mathrm{d} x$

$= - $\frac{\cos $x}{e^{x}$$ } - \int $e^{-x}$ \sin x ; \mathrm{d} x$

To find $\int $e^{-x}$ \sin x ; \mathrm{d} x$ , we use another integration by parts:

$u = \sin x$ , which means that $\mathrm{d} u = \cos x ; \mathrm{d} x$ , and

$\mathrm{d}v= $e^{-x}$; \mathrm{d}x$ , which means that, again, $v= - $$\frac{1}{e^{x}$$ }$ .

$\int $e^{-x}$ \sin x ; \mathrm{d} x$

$=\int u ; \mathrm{d} v$

$= uv - \int v ; \mathrm{d} u$

$= \sin x \cdot \left(- $$\frac{1}{e^{x}$$ } \right) - \int \left(- $$\frac{1}{e^{x}$$ } \right) ; \cdot \cos x ; \mathrm{d} x$

$= - $\frac{ \sin x $}{e^{x}$$ } + \int $\frac{\cos x ; \mathrm{d}$ $x}{e^{x}$ }$

$\int $\frac{\cos x; $\mathrm{d}$x}{e^{x}$} = - $\frac{\cos $x}{e^{x}$$ } - \int $e^{-x}$ \sin x ; \mathrm{d} x$

$\int $\frac{\cos x; $\mathrm{d}$x}{e^{x}$} = - $\frac{\cos $x}{e^{x}$$ } - \left ( - $\frac{ \sin x $}{e^{x}$$ } + \int $\frac{\cos x ; \mathrm{d}$ $x}{e^{x}$ } \right)$

$\int $\frac{\cos x; $\mathrm{d}$x}{e^{x}$} = - $\frac{\cos $x}{e^{x}$$ } + $\frac{ \sin x $}{e^{x}$$ } - \int $\frac{\cos x ; \mathrm{d}$ $x}{e^{x}$ }$

$2 \int $\frac{\cos x; $\mathrm{d}$x}{e^{x}$} = - $\frac{\cos $x}{e^{x}$$ } + $\frac{ \sin x $}{e^{x}$$ }$

$2 \int $\frac{\cos x; $\mathrm{d}$x}{e^{x}$} = $\frac{\sin x -\cos $x}{e^{x}$$ }$

$\int $\frac{\cos x; $\mathrm{d}$x}{e^{x}$} = $\frac{\sin x -\cos $x}{2e^{x}$$ }$

Since

, or,

$- $$\frac{1}{e^{x}$$ } \leq $\frac{\sin x -\cos $x}{2e^{x}$$ } \leq $$\frac{1}{e^{x}$$ }$

for all real , and

$- $\lim_{x\rightarrow \infty }$ $$\frac{1}{e^{x}$$ } = $\lim_{x\rightarrow \infty }$ $$\frac{1}{e^{x}$$ } = 0$ ,

by the Squeeze Theorem,

.

$\int_${0}^{\infty }$ $\frac{\cos x; $\mathrm{d}$x}{e^{x}$}$

$= $\lim_{b\rightarrow \infty }$\int_${0}^{b }$ $\frac{\cos x; $\mathrm{d}$x}{e^{x}$}$

$= $\lim_{ b\rightarrow \infty }$$ $\left.\begin{matrix} $\frac{\sin x -\cos $x}{2e^{x}$$ } \ \textrm{ } \end{matrix}\right|$ $\begin{matrix} b\ \ 0 \end{matrix}$

$= 0 - $\frac{\sin 0 -\cos $0}{2e^{0}$$ }$

$= - $\frac{ 0 -1}{2 \cdot 1 }$ = $\frac{1}{2}$$

Evaluate:

$\int_${0}^{1}$ x \ln x ; \mathrm{d}x$

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First, we will find the indefinite integral, $\int x \ln x ; \mathrm{d}x$ .

We will let $u = \ln x$ and $\mathrm{d}v= x ; \mathrm{d}x$ .

Then,

$\frac{\mathrm{d}$ u }{\mathrm{d} x}= $\frac{1}{x}$ and $\mathrm{d} u = $\frac{\mathrm{d}$ x}{x}$ .

$v= \int x ; \mathrm{d}x =\frac{ x ^{2}$}{2}$

and

$\int x \ln x ; \mathrm{d}x$

$=\int u ; \mathrm{d} v$

$= uv - \int v ; \mathrm{d} u$

$= $\ln{x}$ \cdot $\frac{ x ^{2}$}{2} - \int \left($\frac{ x ^{2}$}{2} \right) $\frac{\mathrm{d}$ x}{x}$

$= $\frac{ x ^{2}$$\ln{x}$}{2} - $\frac{1}{2}$\int x; \mathrm{d} x$

$= $\frac{ x ^{2}$$\ln{x}$}{2} - $\frac{1}{2}$ \cdot $$\frac{x^{2}$$}{2}$

$= $\frac{ x ^{2}$$\ln{x}$}{2} - $$\frac{x^{2}$$}{4}$

Now, this expression evaluated at is equal to

$\frac{ 1 ^{2}$$\ln{1}$}{2} - $$\frac{1^{2}$$}{4} = 0 - $\frac{1}{4}$ = - $\frac{1}{4}$ .

At it is undefined, because $\ln 0$ does not exist.

We can use L'Hospital's rule to find its limit as $x\rightarrow 0$ , as follows:

and , so by L'Hospital's rule,

Therefore,

$\int_${0}^{1}$ x \ln x ; \mathrm{d}x$

$= \left.\begin{matrix} $\frac{ x ^{2}$$\ln{x}$}{2} - $$\frac{x^{2}$$}{4} \ \ \end{matrix}\right| \begin{matrix} 1\ \ 0 \end{matrix}$

$=\lim _{b\rightarrow 0}\left [ \left.\begin{matrix} $\frac{ x ^{2}$$\ln{x}$}{2} - $$\frac{x^{2}$$}{4} \ \ \end{matrix}\right| \begin{matrix} 1\ \ b \end{matrix} \right]$

$= -$\frac{1}{4}$ - 0$

$= -$\frac{1}{4}$$

Evaluate:

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Rewrite the integral as

.

Substitute $u = - $x^{2}$$ . Then

$$\frac{\mathrm{d}$ u}{\mathrm{d} x} = - 2x$ and $-$\frac{1}{2}$ \mathrm{d} u = x ; \mathrm{d} x$ . The lower bound of integration stays $u = - $0^{2}$ = 0$ , and the upper bound becomes , so

$=\frac{1}{2}$ $\lim_{b\rightarrow -\infty }$ \left.\begin{matrix} e ^{u}\ \textrm{ } \ \textrm{ } \end{matrix}\right| \begin{matrix} 0\ \ b \end{matrix}$

Since , the above is equal to

$=\frac{1}{2}$ $(e^{0}$-0) =\frac{1}{2}$ (1) = $\frac{1}{2}$$ .

Evaluate $\int_${1}^{\infty}$$$\frac{4}{x^{5}$$}dx$ .

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By the Formula Rule, we know that $\int_${a}^${\infty}$F(x)dx=\lim_{b\rightarrow \infty}$\int_${a}^{b}$f(x)dx$ . We therefore know that $\int_${1}^{\infty}$$$\frac{4}{x^${5}$$}dx=\lim_{b\rightarrow \infty}$\int_${1}^{b}$$$\frac{4}{x^{5}$$}dx$ .

Continuing the calculation:

$\int_${1}^{b}$$$\frac{4}{x^{5}$$}dx$

$=4\int_${1}^{b}$$$\frac{1}{x^{5}$$}dx$

$=4\int_${1}^{b}$${x^{-5}$}dx$

By the Power Rule for Integrals, $\int $x^{n}$$dx=\frac{x^{n+1}$$}{n+1}+C$ for all $n\neq-1$ with an arbitrary constant of integration . Therefore:

$4\int_${1}^{b}$${x^{-5}$}dx$

.

So,

.

$\int$\frac{x-1}{(x-3)(x+4)}$ dx=$

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In order to evaluate this integral, we will need to use partial fraction decomposition.

$\frac{x-1}{(x-3)(x+4)}$=\frac{A}{x-3}$+\frac{B}{x+4}$

Multiply both sides of the equation by the common denominator, which is

This means that must equal 1, and

$\frac{x-1}{(x-3)(x+4)}$=\frac{2/7}{x-3}$+\frac{5/7}{x+4}$

$\int($\frac{2}{7}$\cdot $\frac{1}{x-3}$+\frac{5}{7}$\cdot $\frac{1}{x+4}$)dx$

$=\int$\frac{2}{7}$\cdot$\frac{1}{x-3}$ dx +\int$\frac{5}{7}$\cdot$\frac{1}{x+4}$dx$

$=\frac{2}{7}$\int$\frac{dx}{x-3}$+\frac{5}{7}$\int$\frac{dx}{x+4}$$

$=\frac{2}{7}$\ln|x-3|+\frac{5}{7}$\ln| x+4|+C$

The answer is $$\frac{2}{7}$\ln|x-3|+\frac{5}{7}$\ln| x+4|+C$ .

Integrate:

$\int $$\frac{dx}{2x^2$+16}$$

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To integrate, we must first make the following substitution:

$u=$\sqrt{2}$x, du=$\sqrt{2}$dx$

Rewriting the integral in terms of u and integrating, we get

$$\frac{1}{\sqrt{2}$}\int $$\frac{du}{u^2$+16}$= $\frac{1}{4\sqrt{2}$}\arctan($\frac{u}{4}$)+C$

The following rule was used to integrate:

$\int $$\frac{du}{u^2$$+a^2$}$=\frac{1}{a}$\arctan($\frac{x}{a}$)+C$

Finally, we replace u with our original x term:

$$\frac{1}{4\sqrt{2}$}\arctan($\frac{\sqrt{2}$x}{4})+C$

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

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