Calculus II — Integrals - Math

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Question

The above graph depicts a function . Does $\lim_{x\rightarrow 1} f(x)$ exist, and why or why not?

Answer

$\lim_{x\rightarrow 1} f(x)$ exists if and only if $\lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$ ;

the actual value of is irrelevant, as is whether is continuous there.

As can be seen,

$\lim_{x\rightarrow 1^{+}} f(x) = 1$ and $\lim_{x\rightarrow 1^{-}} f(x) = 1$ ;

therefore, $\lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$ ,

and $\lim_{x\rightarrow 1} f(x)$ exists.

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Question

The graph depicts a function . Does $\lim_{x\rightarrow 1} g(x)$ exist?

Answer

$\lim_{x\rightarrow 1}g(x)$ exists if and only if $\lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)$ ; the actual value of is irrelevant.

As can be seen, $\lim_{x\rightarrow 1^{+}} g(x) = 1$ and $\lim_{x\rightarrow 1^{-}}g(x) = 1$ ; therefore, $\lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)$ , and $\lim_{x\rightarrow 1} g(x)$ exists.

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Question

The above graph depicts a function . Does $\lim_{x\rightarrow 1} h(x)$ exist, and why or why not?

Answer

$\lim_{x\rightarrow 1} h(x)$ exists if and only if $\lim_{x\rightarrow 1^{-}} h(x) = \lim_{x\rightarrow 1^{+}} h(x)$ . As can be seen from the diagram, $\lim_{x\rightarrow 1^{-}} h(x) = 1$ , but $\lim_{x\rightarrow 1^{+}} h(x) = -1$ . Since $\lim_{x\rightarrow 1^{-}} h(x) \neq \lim_{x\rightarrow 1^{+}} h(x)$ , $\lim_{x\rightarrow 1} h(x)$ does not exist.

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Question

A function is defined by the following piecewise equation:

$f(x)=\left{\begin{matrix} x^{2}+4x-3, x<3\10x-12, x\geq 3 \end{matrix}\right.$

At , the function is:

Answer

The first step to determine continuity at a point is to determine if the function is defined at that point. When we substitute in 3 for , we get 18 as our -value. is thus defined for this function.

The next step is determine if the limit of the function is defined at that point. This means that the left-hand limit must be equal to the right-hand limit at . Substitution reveals the following:

$\lim_{x\rightarrow 3^{-}}f(x)=(3)^{2}+4(3)-3=9+12-3=18$

$\lim_{x\rightarrow 3^{+}}f(x)=10(3)-12=18$

Both sides of the function, therefore, approach a -value of 18.

Finally, we must ensure that the curve is smooth by checking the limit of the derivative of both sides.

$\lim_{x\rightarrow 3^{-}}f^{'}(x)=2(3)+4=10$

$\lim_{x\rightarrow 3^{+}}f^{'}(x)=10$

Since the function passes all three tests, it is continuous.

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Question

Give the $x^{2}$ term of the Maclaurin series of the function $f \left ( x\right ) = \frac{1}{x-2}$

Answer

The $x^{2}$ term of the Maclaurin series of a function has coefficient $\frac{f '' (0) }{2!} = \frac{f '' (0) }{2}$

The second derivative of can be found as follows:

$f \left ( x\right ) = \frac{1}{x-2} = (x-2) ^{-1}$

$f' \left ( x\right ) = -1\cdot (x-2)^{-2} = -1(x-2)^{-2}$

$f'' \left ( x\right ) = (-2) ( -1(x-2)^{-3}) = 2(x-2)^{-3} = \frac{2}{(x-2)^{3}}$

$f'' \left ( 0 \right ) =\frac{2}{(0-2)^{3}} = -\frac{1}{4}$

The coeficient of $x^{2}$ in the Maclaurin series is therefore

$\frac{f '' (0) }{2} = \frac{-\frac{1}{4}}{2} = - \frac{1}{8}$

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Question

Give the term of the Taylor series expansion of the function $f (x) = \log_5 x$ about .

Answer

The $(x-1)^{2}$ term of a Taylor series expansion about is

$\frac{f ' ' (1)}{2!} (x-1)^{2}= \frac{f ' ' (1)}{2}(x-1)^{2}$ .

We can find by differentiating twice in succession:

$f (x) = \log_5 x$

$f'(x) = \frac{1}{x \ln 5}$

$f'(x) =\frac{1}{ \ln 5} \cdot x^{-1}$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{1}{ \ln 5} \cdot x^{-1} \right )$

$f''(x) =\frac{1}{ \ln 5}\cdot \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{-1} \right )$

$f''(x) =\frac{1}{ \ln 5} \cdot \left ( -1 \cdot x^{-2} \right )$

$f''(x) =- \frac{1}{ \ln 5 \cdot x^{2}}$

$f''(1) =- \frac{1}{ \ln 5 \cdot 1^{2}} = - \frac{1}{ \ln 5 }$

so the term is $\frac{f''(1) }{2} ; (x-1)^{2} = - \frac{1}{2 \ln 5 } (x-1)^{2} = - \frac{1}{ \ln 25 } (x-1)^{2}$

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Question

Give the term of the Taylor series expansion of the function $f (x) = \log_5 x$ about .

Answer

The $(x-1)^{2}$ term of a Taylor series expansion about is

$\frac{f ' ' (1)}{2!} (x-1)^{2}= \frac{f ' ' (1)}{2}(x-1)^{2}$ .

We can find by differentiating twice in succession:

$f (x) = \log_5 x$

$f'(x) = \frac{1}{x \ln 5}$

$f'(x) =\frac{1}{ \ln 5} \cdot x^{-1}$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{1}{ \ln 5} \cdot x^{-1} \right )$

$f''(x) =\frac{1}{ \ln 5}\cdot \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{-1} \right )$

$f''(x) =\frac{1}{ \ln 5} \cdot \left ( -1 \cdot x^{-2} \right )$

$f''(x) =- \frac{1}{ \ln 5 \cdot x^{2}}$

$f''(1) =- \frac{1}{ \ln 5 \cdot 1^{2}} = - \frac{1}{ \ln 5 }$

so the term is $\frac{f''(1) }{2} ; (x-1)^{2} = - \frac{1}{2 \ln 5 } (x-1)^{2} = - \frac{1}{ \ln 25 } (x-1)^{2}$

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Question

Give the $x^{2}$ term of the Maclaurin series of the function $f \left ( x\right ) = \frac{1}{x-2}$

Answer

The $x^{2}$ term of the Maclaurin series of a function has coefficient $\frac{f '' (0) }{2!} = \frac{f '' (0) }{2}$

The second derivative of can be found as follows:

$f \left ( x\right ) = \frac{1}{x-2} = (x-2) ^{-1}$

$f' \left ( x\right ) = -1\cdot (x-2)^{-2} = -1(x-2)^{-2}$

$f'' \left ( x\right ) = (-2) ( -1(x-2)^{-3}) = 2(x-2)^{-3} = \frac{2}{(x-2)^{3}}$

$f'' \left ( 0 \right ) =\frac{2}{(0-2)^{3}} = -\frac{1}{4}$

The coeficient of $x^{2}$ in the Maclaurin series is therefore

$\frac{f '' (0) }{2} = \frac{-\frac{1}{4}}{2} = - \frac{1}{8}$

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Question

Find the second derivative of f(x).

$f(x)=\frac{1}{4} e^{2x}-cosx$

Answer

First we should find the first derivative of . Remember the derivative of $e^{2x}$ is $2e^{2x}$ and the derivative of is :

$f'(x)=\frac{1}{2}e^{2x}+sinx$

The second derivative is just the derivative of the first derivative:

$f''(x)=e^{2x}+cosx$

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Question

Find the derivative of the function

.

Answer

We can use the Chain Rule:

Let , so that .

$\frac{df}{dx}=\frac{df}{du}.\frac{du}{dx}=(4u^3-3u^2+2u)\cdot sec^2x=(4tan^3x-3tan^2x+2tanx)\cdot sec^2x$

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Question

Evaluate:

$\dpi{120} \int_{0}^{\frac{\pi }{ 2}}\sin x \sqrt{\cos x }; ; dx$

Answer

Set $u = \cos x$ .

Then $\frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \cos x= -\sin x$ and $\sin x ; dx = - du$ .

Also, since $u = \cos x$ , the limits of integration change to $\cos \frac{\pi }{2} = 0$ and $\cos 0 = 1$ .

Substitute:

$\dpi{120} \int_{0}^{\frac{\pi }{ 2}}\sin x \sqrt{\cos x }; ; dx$

$= \int_{1}^{0} \left (- \sqrt{u} \right ) ; du$

$= \int_{1}^{0} \left (-u ^{\frac{1}{2}} \right ) ; du$

$= \int_{0}^{1} u ^{\frac{1}{2}} ; du$

$\dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \ & \end{matrix}\right|$ $\begin{matrix} 1\ \ 0 \end{matrix}$

$\dpi{150} = \left.\begin{matrix} \ \frac{ 2u \sqrt{u}} {3} & \ & \end{matrix}\right|$ $\begin{matrix} 1\ \ 0 \end{matrix}$

$=\frac{ 2\cdot 1\cdot \sqrt{1}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} = \frac{ 2} {3}$

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Question

Evaluate:

$\dpi{120}\int_{1}^{e^{ 3}}\frac{\sqrt{\ln x}}{x}; ; dx$

Answer

Set $u = \ln x$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and

$du = \frac{dx}{x}$

Also, since $u = \ln x$ , the limits of integration change to $\ln e^{3} = 3$ and $\ln 1= 0$ .

Substitute:

$\dpi{120}\int_{1}^{e^{ 3}}\frac{\sqrt{\ln x}}{x}; ; dx$

$\dpi{120} = \int_{0}^{3}\sqrt{u} ; du$

$= \int_{0}^{3} u ^{\frac{1}{2}} ; du$

$\dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \ & \end{matrix}\right|$ $\begin{matrix} 3\ \ 0 \end{matrix}$

$\dpi{150} = \left.\begin{matrix} \ \frac{ 2u \sqrt{u}} {3} & \ & \end{matrix}\right|$ $\begin{matrix} 3\ \ 0 \end{matrix}$

$=\frac{ 2\cdot 3\cdot \sqrt{3}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} =2\sqrt{3}$

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Question

Determine the indefinite integral:

$\dpi{120} \int\cos x \sqrt{\sin x }; ; dx$

Answer

Set $u = \sin x$ . Then

$\frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \sin x= \cos x$ .

and

$\cos x ; dx = du$

The integral becomes:

$\dpi{120} \int\cos x \sqrt{\sin x }; ; dx$

$\dpi{120} =\int \sqrt{u }; ; du$

$= \int u ^{\frac{1}{2}} ; du$

$\dpi{120} \dpi{150} = \frac{u^{\frac{1}{2}+1}}{{\frac{1}{2}+1}} + C$

$\dpi{150} = \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C$

$\dpi{120} =\frac{ 2u \sqrt{u}} {3} +C$

Substitute back:

$\dpi{120} =\frac{ 2 \sin x \cdot \sqrt{\sin x }} {3} +C$

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Question

Determine the indefinite integral:

$\dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}; ; dx$

Answer

$\log_{10} x = \frac{\ln x}{\ln 10}$ , so this can be rewritten as

$\dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}; ; dx$

$\dpi{120}=\int\frac{\sqrt{\ln x}}{x \sqrt{\ln 10} }; ; dx$

$\dpi{120}= \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }; ; dx$

Set $u = \ln x$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and

$du = \frac{dx}{x}$

Substitute:

$\dpi{120} \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }; ; dx$

$\dpi{120} =\frac{1}{\sqrt{\ln 10}} \int\sqrt{u} ; du$

$\dpi{120} =\frac{1}{\sqrt{\ln 10}} \int u ^{\frac{1}{2}} ; du$

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C \right )$

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2u \sqrt{u}} {3} + C \right )$

The outer factor can be absorbed into the constant, and we can substitute back:

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2 \ln x \cdot \sqrt{ \ln x}} {3} \right )+ C$

$= \frac{ 2 \sqrt{ \ln x} \cdot \ln x} {3 \cdot \sqrt{\ln 10}} + C$

$= \frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$

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Question

$\int_{4}^{9}\frac{\sqrt{x}}{1-x}dx=$

Answer

This integral will require a u-substitution.

Let $u=\sqrt{x}$ .

Then, differentiating both sides, $du=\frac{1}{2\sqrt{x}}dx$ .

We need to solve for dx in order to replace all x terms with u terms.

$dx=(2\sqrt{x})du$ .

$\int\frac{\sqrt{x}}{1-x}\cdot(2\sqrt{x})du=\int\frac{2xdu}{1-x}$

This is a little tricky because we stilll have x and u terms mixed together. We need to go back to our original substitution.

$u=\sqrt{x}\Rightarrow x=u^2$

$\int\frac{2xdu}{1-x}=\int\frac{2u^2 du}{1-u^2}$

Now we have an integral that looks more manageable. First, however, we can't forget about the bounds of the definite integral. We were asked to evaluate the integral from to . Because $u=\sqrt{x}$ , the bounds will change to $\sqrt{4}=2$ and $\sqrt{9}=3$ .

Essentially, we have made the following transformation:

$\int_{4}^{9}\frac{\sqrt{x}}{1-x}dx=\int_{2}^{3}\frac{2u^2}{1-u^2}du$ .

The latter integral is easier to evaluate.

$\int_{2}^{3}\frac{2u^2}{1-u^2}du=-2\int_{2}^{3}\frac{-u^2}{1-u^2}du$

$=-2\int_{2}^{3}\frac{1-u^2-1}{1-u^2}du=-2\int_{2}^{3}\left ( \frac{1-u^2}{1-u^2}+\frac{-1}{1-u^2} \right )du$

$=-2\int_{2}^{3}\left ( 1-\frac{1}{1-u^2}\right )du$

At this point, we can separate the integral into two smaller integrals.

$-2\int_{2}^{3}\left ( 1-\frac{1}{1-u^2}\right )du=-2\int_{2}^{3}1\cdot du+2\int_{2}^{3}\frac{1}{1-u^2}du$ .

The integral $-2\int_{2}^{3}1\cdot du$ evaluates to -2, so now we just need to worry about the other integral. This will require the use of partial fraction decomposition. We need to rewrite $\frac{1}{1-u^2}$ as the sum of two fractions.

$\frac{1}{1-u^2}=\frac{1}{(1-u)(1+u)}=\frac{A}{1-u}+\frac{B}{1+u}$

We need to solve for the values of A and B.

$\frac{A}{1-u}+\frac{B}{1+u}=\frac{1}{(1-u)(1+u)}$

This means that and . This is a relatively simple system of equations to solve, so I won't go into detail. The end result is that $A=B=\frac{1}2{}$ .

$\frac{1}{1-u^2}=\frac{\frac{1}{2}}{1-u}+\frac{\frac{1}{2}}{1+u}$

Let's now go back to the integral $2\int_{2}^{3}\frac{1}{1-u^2}du$ .

$2\int_{2}^{3}\frac{1}{1-u^2}du=2\int_{2}^{3}\left ( \frac{\frac{1}{2}}{1-u}+\frac{\frac{1}{2}}{1+u} \right )du$

Distribute the 2 to both integrals and separate it into two integrals.

$=\int_{2}^{3}\frac{1}{1-u}du + \int_{2}^{3}\frac{1}{1+u}du$

$=-\ln|1-u| ]_{2}^{3} + \ln|1+u|]_{2}^{3}$

$=-\ln|-2|+\ln|-1|+\ln4-\ln3$

$=-\ln2+\ln1+\ln4-\ln3$

$=\ln\frac{2}{3}$ .

Remember we need to add this value back to the value of $-2\int_{2}^{3}1\cdot du$ , which we already determined to be -2.

The final answer is $-2+\ln\frac{2}{3}$ .

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Question

$\int \frac{3}{x}=$

Answer

The integral of $\frac{1}{x}$ is $\ln \left | x \right |+C$ . The constant 3 is simply multiplied by the integral.

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Question

Evaluate the integral below:

$f(x)=\int \frac{dx}{x^2-16}$

Answer

In this case we have a rational function as $\frac{N(x)}{D(x)}$ , where

and

$D(x)=\frac{1}{x^2-16}$

can be written as a product of linear factors:

$\frac{1}{x^2-16}=\frac{1}{(x-4)(x+4)}\equiv \frac{A}{x-4}+\frac{B}{x+4}$

It is assumed that A and B are certain constants to be evaluated. Denominators can be cleared by multiplying both sides by (x - 4)(x + 4). So we get:

First we substitute x = -4 into the produced equation:

$1=-8B\Rightarrow B=-\frac{1}{8}$

Then we substitute x = 4 into the equation:

$1=8A\Rightarrow A=\frac{1}{8}$

Thus:

$\frac{1}{(x-4)(x+4)}=\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4}$

Hence:

$\int \frac{dx}{x^2-16}=\int(\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4})dx$

$=\frac{1}{8}ln\left | x-4 \right |-\frac{1}{8}ln\left |x+4 \right |+c$

$=\frac{1}{8}(ln\left | x-4 \right |-ln\left |x+4 \right |)+c$

$=\frac{1}{8}ln\left | \frac{x-4}{x+4}\right |+c$

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Question

$\int cos(x)\sin(x){\mathrm{d} x}=?$

Answer

To integrate $\cos(x)\sin(x)$ , we need to get the two equations in terms of each other. We are going to use "u-substitution" to create a new variable, , which will equal $\cos(x)$ .

Now, if $u=\cos(x)$ , then

$\frac{\mathrm{d}u }{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-sin(x)$

Multiply both sides by $\mathrm d {x}$ to get the more familiar:

${\mathrm{d} u}=-\sin(x)\mathrm{d}x$

Note that our $\mathrm d{u}=-\sin(x)$ , and our original equation was asking for a positive $\sin(x)$ .

That means if we want $\int\cos(x)\sin(x)$ in terms of , it looks like this:

$\int u\cdot (-\mathrm d{u}})$

Bring the negative sign to the outside:

$-\int u\mathrm d{u}$ .

We can use the power rule to find the integral of :

$-(\frac{1}{2}u^2+c)$

Since we said that $u=\cos(x)$ , we can plug that back into the equation to get our answer:

$-(\frac{1}{2}\cos^2(x)+c)$

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Question

What is the anti-derivative of ?

Answer

To find the indefinite integral of our expression, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

First we need to realize that . From there we can solve:

$\int2x{\mathrm{d} x}=\frac{2x^{1+1}}{1+1}+c$

When taking an integral, be sure to include a at the end of everything. stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being or instead of just .

$\int2x{\mathrm{d} x}=\frac{2x^{2}}{2}+c$

$\int2x{\mathrm{d} x}=x^2+c$

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Question

What is the indefinite integral of ?

Answer

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

Remember that, when taking the integral, we treat constants as that number times $x^{0}$ since anything to the zero power is . For example, treat as .

$\int x^3+2{\mathrm{d} x}=\frac{x^{3+1}}{3+1}+\frac{2x^{0+1}}{0+1}+c$

When taking an integral, be sure to include a at the end of everything. stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being or instead of just .

$\int x^3+2{\mathrm{d} x}=\frac{x^{4}}{4}+\frac{2x^{1}}{1}+c$

$\int x^3+2{\mathrm{d} x}=\frac{1}{4}x^4+2x+c$

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