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

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Question

Evaluate :

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

Answer

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, $f'(0) = \sin \sqrt{0^2 + \frac{\pi ^2}{4}} =1$ .

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Question

Find the result:

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

Answer

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

$e^x \cdot u \ln u$

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

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

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Question

Suppose we have the function

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

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

Answer

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

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

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Question

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

Answer

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 $f'(4)=2(4)^{2}+5(4)-4=32+20-4=52-4=48$ .

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Question

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

Answer

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, $f'(1)=6(1)^{2}-3(1)+10=6-3+10=3+10=13$ .

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Question

Given

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

Answer

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,

$f'(6)=\frac{1}{6^{2}}-2(6)+3=\frac{1}{36}-12+3=\frac{1}{36}-9=\frac{1}{36}-\frac{324}{36}=-\frac{323}{36}$

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Question

Given

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

Answer

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,

$f'(2)=\frac{2}{2^{2}+4(2)-5}=\frac{2}{4+8-5}=\frac{2}{4+3}=\frac{2}{7}$ .

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Question

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

Answer

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

Use the fundamental theorem of calculus to evaluate:

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

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

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Question

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

Answer

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

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

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

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

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Question

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

Answer

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

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

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

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

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Question

Evaluate the following integral

$\int_0^5 3x^4-12x^3+10x+12dx$

Answer

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:

$\frac{3x^5}{5}-3x^4+5x^2+12x+c\mid_0^5$

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)

$\frac{3(5)^5}{5}-3(5)^4+5(5)^2+12(5)+c=1875-1875+125+60+c=185+c$

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.

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Question

Evaluate:

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

Answer

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

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

$\lim_{x\rightarrow \infty } \frac{\sin x -\cos x}{2e^{x} } = 0$ .

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

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

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

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

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

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Question

Evaluate:

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

Answer

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:

$\lim_{x\rightarrow 0}\frac{2 x ^{2}\ln{x} - x^{2}}{4} =\lim_{x\rightarrow 0} \frac{2 \ln{x} - 1}{\frac{4}{x ^{2}}}$

$\lim_{x\rightarrow 0} \left (2 \ln{x} - 1 \right ) = - \infty$ and $\lim_{x\rightarrow 0} \frac{4}{x ^{2}} = \infty$ , so by L'Hospital's rule,

$\lim_{x\rightarrow 0} \frac{2 \ln{x} - 1}{\frac{4}{x ^{2}}}=\lim_{x\rightarrow 0} \frac{\frac {2}{x} }{\frac{-8}{x ^{3}}} = \lim_{x\rightarrow 0}\left ( -\frac{x^{2}}{4} \right ) = 0$

Therefore,

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

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

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

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

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

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Question

Evaluate:

$\int_{0 }^{\infty} \frac{x; \mathrm{d}x}{e^{x^{2}}}$

Answer

Rewrite the integral as

$\int_{0 }^{\infty} \frac{x; \mathrm{d}x}{e^{x^{2}}} = \int_{0 }^{\infty}e^{-x^{2}}\cdot x ; \mathrm{d}x} }$ .

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 $\lim_{x\rightarrow \infty } - x^{2} = - \infty$ , so

$\int_{0 }^{\infty} \frac{x; \mathrm{d}x}{e^{x^{2}}} = \int_{0 }^{\infty}e^{-x^{2}}\cdot x ; \mathrm{d}x} }=-\frac{1}{2} \int_{0 }^{-\infty}e^{u} \mathrm{d}u}= \frac{1}{2} \int_{-\infty }^{0}e^{u} \mathrm{d}u}$

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

Since $\lim_{b\rightarrow -\infty } e ^{u} = 0$ , the above is equal to

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

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Question

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

Answer

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 with an arbitrary constant of integration . Therefore:

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

$=4({\frac{x^{-5+1}}{-5+1}})_{1}^{b}$

$=4(\frac{x^{-4}}{-4}})_{1}^{b}$

$=4(-{\frac{1}{4x^{4}}})_{1}^b$ .

So, $\lim_{b\rightarrow \infty }4(-{\frac{1}{4x^{4}}})_{1}^b$

$=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}-(-\frac{1}{4}})]$

$=\lim_{b\rightarrow \infty }4[-{\frac{1}{4b^{4}}+\frac{1}{4}})]$

$=\lim_{b\rightarrow \infty }[-{\frac{1}{b^{4}}+1}]$ .

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Question

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

Answer

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

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Question

Integrate:

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

Answer

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:

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

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Question

Evaluate :

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

Answer

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, $f'(0) = \sin \sqrt{0^2 + \frac{\pi ^2}{4}} =1$ .

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Question

Find the result:

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

Answer

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

$e^x \cdot u \ln u$

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

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

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Question

Suppose we have the function

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

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

Answer

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

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

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