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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #171 : Equations

Evaluate the following integral:

$\int( \csc(5x)+x)dx$

Possible Answers:

$\frac{\ln\left | \csc(5x)+\cot(5x)\right |}{5}+\frac{x^2}{2}+C$

$\frac{\ln\left | \csc(5x)-\cot(5x)\right |}{5}+{x^2}+C$

$\frac{\ln\left | \csc(5x)-\cot(5x)\right |}{5}+\frac{x^2}{2}+C$

${\ln\left | \csc(5x)-\cot(5x)\right |}+\frac{x^2}{2}+C$

Correct answer:

$\frac{\ln\left | \csc(5x)-\cot(5x)\right |}{5}+\frac{x^2}{2}+C$

Explanation:

To evaluate this integral, it is easier to split it into two integrals:

$\int \csc(5x)dx+ \int x dx$

For the first integral, the following substitution must be made:

u=5x, du=5dx

Rewriting the integral and integrating, we get:

$\frac{1}{5}\int \csc(u)du=\frac{1}{5}\ln\left | \csc(u)-\cot(u)\right |+C$

The following rule was used to integrate:

$\int \csc(x)dx=\ln\left | \csc(x)-\cot(x) \right |+C$

Replacing u with the original term, we get

$\frac{1}{5}\ln\left | \csc(5x)-\cot(5x)\right |+C$

Next, we evaluate the second integral:

$\int xdx=\frac{x^2}{2}+C$

which was found using the following rule:

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

Combining the results - and the constants of integration - we get

$\frac{\ln\left | \csc(5x)-\cot(5x)\right |}{5}+\frac{x^2}{2}+C$

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Example Question #2252 : Calculus

Set up, but don't evaluate an integral expression which will find the area of the region bound by the lines x=5 , x=25 , the x-axis, and the function v(x)

v(x)=14x^6+5x^4-3x^2+4

Possible Answers:

$\int_{5}^{25}14x^6+5x^4-3x^2+4dx$

$\int_{0}^{5}14x^6+5x^4-3x^2+4dx$

$\int_{0}^{20}14x^6+5x^4-3x^2+4dx$

$\int_{25}^{25}14x^6+5x^4-3x^2dx$

Correct answer:

$\int_{5}^{25}14x^6+5x^4-3x^2+4dx$

Explanation:

Set up, but don't evaluate an integral expression which will find the area of the region bound by the lines x=5 , x=25 , the x-axis, and the function v(x)

v(x)=14x^6+5x^4-3x^2+4

First, we need the correct limits of integration. We want 5 as the lower limit and 25 as the upper limit.

$\int_{5}^{25}$

Next, add in our function and the dx and we are all set!

$\int_{5}^{25}14x^6+5x^4-3x^2+4dx$

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Example Question #171 : Integral Expressions

Set up, but do not evaluate, an integral expression for the area of the region bound by the y-axis, h(t), and the lines ,

h(t)=5e^t+18t^2-13

Possible Answers:

$\int_{0}^{6}5e^t+18t^2-13dt$

$\int_{0}^{5}5e^t+18t^2-13dt$

$\int_{5}^{6}5e^t+18t^2-13dt$

$\int_{5}^{6}5e^t+18t^2-13$

Correct answer:

$\int_{5}^{6}5e^t+18t^2-13dt$

Explanation:

Set up, but do not evaluate, an integral expression for the area of the region bound by the y-axis, h(t), and the lines ,

h(t)=5e^t+18t^2-13

To set up a definite integral, we first need our limits of integration:

In this case, they will be 5 and 6, because those are the lines which are the boundaries of our region.

$\int_{5}^{6}$

Finally, we want to include our function

$\int_{5}^{6}5e^t+18t^2-13dt$

Don't forget the dt, so that we know which variable we would integrate with respect to.

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Example Question #171 : Integral Expressions

Evaluate the following integral:

$\int (t^2+10\csc(t)+e^{3t})dt$

Possible Answers:

$\frac{t^3}{3}+10\ln\left | \csc(t)-\cot(t)\right |+{e^{3t}}+C$

$\frac{t^3}{3}+10\ln\left | \csc(t)-\cot(t)\right |+\frac{e^{3t}}{3}+C$

$\frac{t^3}{3}+10\left | \csc(t)-\cot(t)\right |+\frac{e^{3t}}{3}+C$

$\frac{t^3}{3}+10\ln\left | \csc(t)+\cot(t)\right |+\frac{e^{3t}}{3}+C$

$\frac{t^3}{3}+\ln\left | \csc(t)-\cot(t)\right |+\frac{e^{3t}}{3}+C$

Correct answer:

$\frac{t^3}{3}+10\ln\left | \csc(t)-\cot(t)\right |+\frac{e^{3t}}{3}+C$

Explanation:

The integral is equal to

$\frac{t^3}{3}+10\ln\left | \csc(t)-\cot(t)\right |+\frac{e^{3t}}{3}+C$

and was found using the following rules:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int \csc(x)dx=\ln\left | \csc(x)-\cot(x)\right |+C$ , $\int e^{ax}dx=\frac{e^{ax}}{a}+C$

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Example Question #2255 : Calculus

Find the indefenite integral $\int \frac{sin(ln(x))}{x}dx$

Possible Answers:

$\frac{-cos(ln(x))}{x^2}$

cos(ln(x))+C

sin(ln(x))+cos(ln(x))+c

-cos(ln(x))+C

$\frac{sin(ln(x))}{x^2}+C$

Correct answer:

-cos(ln(x))+C

Explanation:

We need to use substitution to solve. Let

u=ln(x)

Differentiating, gives

$du=\frac{1}{x}dx$

Solving for dx gives

dx=xdu

Plugging into the original integral gives

$\int \frac{sin(ln(x))}{x}dx=\int sin(ln(x))\cdot \frac{1}{x}dx=\int sin(u)\cdot \frac{1}{x}xdu$

Simplifying gives

$\int sin(u)du=-cos(u)+C$

Plugging back in for u gives

-cos(ln(x))+C

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Example Question #171 : Writing Equations

Find the indefenite integral $\int x^2(x^3+3)^8dx$

Possible Answers:

$\frac{x^3(x^4+3)^9}{36}$

$\frac{x^3(x^4+3)^9}{108}$

$\frac{x^3(x^4+3)^8}{12}$

$\frac{(x^3+3)^9}{27}$

$\frac{x^3(x^3+3)^9}{81}$

Correct answer:

$\frac{(x^3+3)^9}{27}$

Explanation:

We need to use substitution to solve. Let

u=(x^3+3)^8

Differentiating gives

du=3x^2dx

Solving for dx, so that we can plug back in, gives

$dx=\frac{du}{3x^2}$

Plugging in gives

$\int x^2(x^3+3)^8dx=\int x^2 \cdot u^8 \cdot\frac{du}{3x^2}$

Simplifying gives

$\int u^8 \cdot\frac{du}{3}=\frac{1}{3}\int u^8du$

Integrating gives

$\frac{1}{3} \cdot \frac{u^9}{9}+C=\frac{u^9}{27}+C$

Plugging u back in gives

$\frac{(x^3+3)^9}{27}$

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Example Question #2257 : Calculus

Find the indefenite integral $\int xsin(x)dx$

Possible Answers:

sin(x)-cos(x)+C

$\frac{x^2}{2}sin(x)+C$

cos(x)-sin(x)+C

$\frac{x^2cos(x)}{2}+C$

-xcos(x)+sin(x)+C

Correct answer:

-xcos(x)+sin(x)+C

Explanation:

We need to use integration by parts, which says

$\int udv=uv-\int vdu$

Let

u=x, dv=sin(x)dx

Differentiating u and integrating dv, gives

du=dx, v=-cos(x)

Plugging into the equation, gives

$\int xsin(x)dx=x \cdot (-cos(x))-\int (-cos(x))dx$

$=-xcos(x)+\int cos(x)dx$

Now, we can integrate the last term to get

=-xcos(x)+sin(x)+C

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Example Question #172 : Integral Expressions

Find the indefenite integral $\int x^2ln(x)dx$

Possible Answers:

$y'=\frac{1}{x^2}+C$

$y'=\frac{x^3}{ln(x)}+C$

$\frac{x^3}{3}(ln(x)-\frac{1}{3})+C$

y'=x^2+C

$y'=\frac{x^2}{3}+C$

Correct answer:

$\frac{x^3}{3}(ln(x)-\frac{1}{3})+C$

Explanation:

We need to use integration by parts, which says

$\int udv=uv-\int vdu$

Let

u=ln(x), dv=x^2dx

Differentiating u and integrating dv, gives

$du=\frac{dx}{x}, v=\frac{x^3}{3}$

Plugging into the equation, gives

$\int x^2ln(x)dx=ln(x)\frac{x^3}{3}-\int \frac{x^3}{3}\frac{dx}{x}$

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

Now, we can integrate the last term to get

$=\frac{x^3}{3}ln(x)-\frac{1}{3}\cdot \frac{x^3}{3}+C$

$=\frac{x^3}{3}ln(x)-\frac{x^3}{9}+C$

$=\frac{x^3}{3}(ln(x)-\frac{1}{3})+C$

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Example Question #1224 : Functions

Solve the integral:

$\int e^{2s}\sec^2(e^{2s})ds$

Possible Answers:

$\tan(e^{2s})+C$

$\frac{1}{2}\ln\left |\sec(e^{2s})+\tan(e^{2s}) \right |+C$

$\frac{1}{2}\tan(e^{2s})+C$

$-\frac{1}{2}\tan(e^{2s})+C$

Correct answer:

$\frac{1}{2}\tan(e^{2s})+C$

Explanation:

To solve the integral, we must first make the following substitution:

$u=e^{2s}$

$du=2e^{2s}ds$

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u = e^u \frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, rewrite the integral and integrate:

$\frac{1}{2}\int \sec^2(u)du=\frac{1}{2}\tan(u)+C$

The integral was found using the rule

$\int \sec^2(x)dx=\tan(x)+C$

Finally, replace u with the original term:

$\frac{1}{2}\tan(e^{2s})+C$ .

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Example Question #2260 : Calculus

A very important physics formula is called the continuity equation. We will only consider the 1-dimensional continuity equation for now.

In the continuity equation, we're given that:

$\int \frac{\partial u}{\partial t} dx=Flux_{in}- Flux_{out}$ , where is a function of and .

Rewrite the right side of the equation in integral form.

Possible Answers:

$\int_{out}^{in} (\frac{d}{dx} Flux ) dx$

$\int_{out}^{in} (Flux ) dx$

$\int_{in}^{out} (Flux ) dx$

$\int_{in}^{out} (\frac{d}{dx} Flux ) dx$

Correct answer:

$\int_{out}^{in} (\frac{d}{dx} Flux ) dx$

Explanation:

Recall that the fundamental theorem of calculus states that:

$F(b)-F(a)=\int_{a}^{b}\frac{d}{dx}F(x) dx$

Using this knowledge we write the right side as:

$Flux_{in}- Flux_{out}=$ $\int_{out}^{in} (\frac{d}{dx} Flux ) dx$

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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #171 : Equations

Example Question #2252 : Calculus

Example Question #171 : Integral Expressions

Example Question #171 : Integral Expressions

Example Question #2255 : Calculus

Example Question #171 : Writing Equations

Example Question #2257 : Calculus

Example Question #172 : Integral Expressions

Example Question #1224 : Functions

Example Question #2260 : Calculus

All Calculus 1 Resources

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