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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Indefinite Integrals

Evaluate the following integral:

$\int (x^6 + \frac{1}{\sqrt{4x^2-1}})dx$

Possible Answers:

$\frac{x^7}{7}-\frac{\arcsin(2x)}{2}+C$

$\frac{x^7}{7}+\frac{\arcsin(2x)}{2}+C$

$\frac{x^7}{7}+\frac{\arcsin(x)}{2}+C$

$\frac{x^7}{7}+\frac{\arcsin(2x)}{2}$

${x^7}+\frac{\arcsin(2x)}{2}+C$

Correct answer:

$\frac{x^7}{7}+\frac{\arcsin(2x)}{2}+C$

Explanation:

To evaluate the integral, we must sum two integrals:

$\int x^6 dx + \int \frac{dx}{\sqrt{4x^2-1}}$

The first integral is equal to

$\frac{x^7}{7}+C$

and was found using the following rule:

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

The second integral is found by first making a subsitution:

u=2x, du=2dx

Now, rewrite the integral and solve:

$\frac{1}{2}\int \frac{du}{\sqrt{u^2-1}}=\frac{1}{2}\arcsin(u)+C$

We used the following rule to integrate:

$\int \frac{dx}{\sqrt{x^2-a^2}}=\arcsin(\frac{x}{a})+C$

Finally, replace with our original term and add the results of the integrations together:

$\frac{x^7}{7}+\frac{\arcsin(2x)}{2}+C$

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Example Question #291 : Finding Integrals

Evaluate the following integral:

$\int (\sec(x)\tan(x)+xe^{5x})dx$

Possible Answers:

$\tan(x)+\frac{xe^{5x}}{5}+\frac{e^{5x}}{5}+C$

$\sec(x)+{xe^{5x}}+\frac{e^{5x}}{5}+C$

$\sec(x)+\frac{xe^{5x}}{5}+\frac{e^{5x}}{5}+C$

$\sec(x)+\frac{xe^{5x}}{5}+\frac{e^{5x}}{25}+C$

Correct answer:

$\sec(x)+\frac{xe^{5x}}{5}+\frac{e^{5x}}{25}+C$

Explanation:

To integrate, we must rewrite the integral into the sum of two integrals:

$\int \sec(x)\tan(x)dx+\int xe^{5x}dx$

The first integral is equal to

$\sec(x)+C$

and was found using the rule identical to the integral itself.

The second integral must be solved using integration by parts, which is given by

$\int udv = uv-\int vdu$

We are going to let

$u=x, dv=e^{5x}$

which means

$du=dx, v=\frac{e^{5x}}{5}$

The derivative and integral were found using the following rules:

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

Now, use the above formula and integrate:

$\frac{xe^{5x}}{5}+\int \frac{e^{5x}}{5}dx=\frac{xe^{5x}}{5}+\frac{e^{5x}}{25}+C$

The integral was performed using the same rule as above.

Putting everything together, we get our final answer:

$\sec(x)+\frac{xe^{5x}}{5}+\frac{e^{5x}}{25}+C$

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Example Question #641 : Integrals

Evaluate the following integral:

$\int (8x+xe^{4x^2})dx$

Possible Answers:

$x^2+\frac{e^{4x^2}}{4}+C$

$4x^2+\frac{e^{4x^2}}{8}+C$

$4x^2-\frac{e^{4x^2}}{8}+C$

$4x^2+e^{4x^2}+C$

Correct answer:

$4x^2+\frac{e^{4x^2}}{8}+C$

Explanation:

The integration of the function is easiest performed when you sum two separate integrals:

$\int 8xdx+\int xe^{4x^2}dx$

The first integral is equal to

4x^2+C

and was found using the following rule:

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

The second integration is done by first making a substitution:

u=4x^2, du=8xdx

Now, rewrite the second integral and solve:

$\frac{1}{8} \int e^u du=\frac{1}{8} e^u +C$

The integral was found using the following rule:

$\int e^x dx=e^x+C$

Now, replace u with the original term and add the two integrals together to get

$4x^2+\frac{e^{4x^2}}{8}+C$

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Example Question #17 : Indefinite Integrals

Evaluate $\int \frac{2dx}{3x^{\frac{1}{3}}(4 + x^{\frac{4}{3}})}$ .

Possible Answers:

$\ln \left | 4 + x^{\frac{4}{3}} \right | +C$

$\frac{1}{2} \tan^{-1} (\frac{x^{\frac{2}{3}}}{2}) + C$

$2 \tan^{-1} (\frac{x^{\frac{2}{3}}}{2}) + C$

$\tan^{-1} (\frac{x^{\frac{2}{3}}}{2}) +C$

Correct answer:

$\frac{1}{2} \tan^{-1} (\frac{x^{\frac{2}{3}}}{2}) + C$

Explanation:

This follows the basic intergral form for the Inverse Tangent. The form is

$\int \frac{du}{a^{2} + u^{2}} = \frac{1}{a} \tan^{-1} (\frac{u}{a}) + C$ ,where is a constant.

If you don't immediately see how $\int \frac{2dx}{3x^{\frac{1}{3}}(4 + x^{\frac{4}{3}})}$ matches the Inverse Tangent integral, then rewrite so that only the $(4 + x^{\frac{4}{3}})$ is in the denominator.

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

Then rewrite the as 2^2 ,and the $x^{\frac{4}{3}}$ as $(x^{\frac{2}{3}})^2$ , so they match the a^2 and the u^2 in the integral form.

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

This matches the Inverse Tangent integral exactly with "u", "a", and "du" as follows

$u = x^{\frac{2}{3}}$

$du=\frac{2}{3} x^{\frac{-1}{3}} dx$

a = 2

Now from the basic integral form, we can simply plug in "u" and "a".

$\int \frac{du}{{\color{Green} a}^{2} + {\color{Blue} u}^{2}} = \frac{1}{{\color{Green} a}} \tan^{-1} (\frac{{\color{Blue} u}}{{\color{Green} a}}) + C$

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

This gives the correct answer of

$\frac{1}{2} \tan^{-1} (\frac{x^{\frac{2}{3}}}{2}) + C$

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Example Question #11 : Indefinite Integrals

Evaluate the following integral:

$\int 5x\cos(x)dx$

Possible Answers:

$\frac{5x^2}{2}\sin(x)+C$

$5x\sin(x)+5\cos(x)+C$

$5x\sin(x)-5\cos(x)+C$

$5x\sin(x)+\cos(x)+C$

Correct answer:

$5x\sin(x)+5\cos(x)+C$

Explanation:

To evaluate the integral, we must integrate by parts:

$\int udv=uv-\int vdu$

So, we must assign and :

$u=5x, dv=\cos(x)dx$

Next, we derive and integrate:

$du=5dx, v=\sin(x)$

The derivative and integral were performed using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\int \cos(x)dx=\sin(x)+C$

Next, use the above formula and integrate:

$5x\sin(x) - \int 5\sin(x)dx= 5x\sin(x)+5\cos(x)+C$

The following rule was used to integrate:

$\int \sin(x)dx=-\cos(x)+C$

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Example Question #11 : Indefinite Integrals

Calculate the indefinite integral

$\int \cos(x)\, dx$

Possible Answers:

$\sin(x)+C$

$\csc(x)+C$

$\tan(x)+C$

$\sec(x)+C$

Correct answer:

$\sin(x)+C$

Explanation:

In order to calculate the indefinite integral, we apply the trigonometric property which states

$\int \cos(x)\,dx = \sin(x)$

As such, the indefinite integral is

$\sin(x)+c$

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Example Question #12 : Indefinite Integrals

Calculate the indefinite integral

$\int 2x\, dx$

Possible Answers:

x^2+C

x^2

$\frac{x^2}{2}$

$\frac{x^2}{2}+C$

Correct answer:

x^2+C

Explanation:

In order to calculate the indefinite integral, we apply the inverse power rule which states

$\int x^n\,dx = \frac{x^{n+1}}{n+1}$

Applying this to the problem in this question we get

$\int 2x\,dx = \frac{2x^2}{2}+c$

As such the indefinite integral is

x^2+C

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Example Question #21 : Indefinite Integrals

Solve: $\int\:3x^2+ln(x) \: dx$

Possible Answers:

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

x^3-xln(x)-x+C

x^3+xln(x)+x+C

x^3+xln(x)-x+C

$x^3+\frac{1}{x}+C$

Correct answer:

x^3+xln(x)-x+C

Explanation:

The indefinite integral can be split into two separate integrals.

$\int\:3x^2+ln(x) \: dx = \int\:3x^2\:dx+\int ln(x) \: dx$

Evaluate each integral.

$\int\:3x^2\:dx = \frac{3x^{2+1}}{2+1} +C_1 = x^3+C_1$

The answer to the first integral is: x^3+C_1

Be careful since the answer to $\int ln(x) \: dx$ is not $\frac{1}{x}$ as this is the derivative of ln(x) . In order to solve this integral, we will need to use integration by parts.

If we let u=ln(x) , we will have $du= \frac{1}{x} dx$ by differentiation, and if we let dv=dx , we will have v=x by integration. The constant can be added in at the end of the problem.

Write the formula for integration by parts.

$\int u\:dv = uv-\int v\:du$

Substitute the terms into the formula.

$\int u\:dv =ln(x)\cdot x-\int(x\cdot \frac{1}{x}\:dx)$

Simplify the terms inside the integral and evaluate.

$\int u\:dv =xln(x)-\int\:dx$

$\int u\:dv =xln(x)-x+C_2$

The answer to the second integral is: xln(x)-x+C_2

Combine the two answers. The constants C_1, C_2 can be combined to be a single constant term at the end.

The answer is: x^3+xln(x)-x+C

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Example Question #21 : Indefinite Integrals

Evaluate: $\int 2x\sqrt{x-1}\:dx$

Possible Answers:

$\frac{21}{15}(x-1)^{\frac{4}{3}}+\frac{16}{3}(x-1)^{\frac{3}{2}} +C$

$\frac{2}{5}(x-1)^{\frac{1}{3}}+\frac{3}{4}(x-1)^{\frac{5}{2}} +C$

$\frac{4}{5}(x-1)^{\frac{5}{2}}-\frac{4}{3}(x-1)^{\frac{3}{2}} +C$

$\frac{4}{3}x(x-1)^{\frac{3}{2}}+C$

$\frac{4}{5}(x-1)^{\frac{5}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}} +C$

Correct answer:

$\frac{4}{5}(x-1)^{\frac{5}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}} +C$

Explanation:

This integral will require substitution with both and terms.

If we let u=x-1 , then x =u+1 .

Differentiate with respect to .

du=dx

Substitute all terms back into the integral.

$\int 2x\sqrt{x-1}\:dx = \int 2(u+1)\sqrt{u}\:du$

The $\sqrt u$ term is also the same as $u^{\frac{1}{2}}$ , which can be multiplied within the terms of the parentheses. Simplify the integral.

$\int 2(u+1)\sqrt{u}\:du = \int 2u^{\frac{3}{2}}+2u^{\frac{1}{2}}\: du$

Evaluate this integral.

$\int 2u^{\frac{3}{2}}+2u^{\frac{1}{2}}\: du = 2[u^{\frac{5}{2}}\cdot \frac{2}{5}] + 2[u^{\frac{3}{2}}\cdot \frac{2}{3}] =\frac{4}{5}u^{\frac{5}{2}}+\frac{4}{3}u^{\frac{3}{2}}+C$

Resubstitute .

$\frac{4}{5}u^{\frac{5}{2}}+\frac{4}{3}u^{\frac{3}{2}} +C= \frac{4}{5}(x-1)^{\frac{5}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}} +C$

The answer is: $\frac{4}{5}(x-1)^{\frac{5}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}} +C$

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Example Question #23 : Indefinite Integrals

Evaluate: $\int \frac{6}{x^2+4x+10}\:dx$

Possible Answers:

$\frac{\sqrt6}{6} tan^{-1}(\frac{x+2}{\sqrt6})+C$

$\sqrt6\:sec^{-1}(\frac{x+2}{\sqrt6})+C$

$\sqrt6\:tan^{-1}(\frac{x+2}{\sqrt6})+C$

$\frac{\sqrt{6}\: ln\left | x^2+4x+10\right |}{2x+4}+C$

$\frac{6ln\left | x^2+4x+10\right |}{2x+4}+C$

Correct answer:

$\sqrt6\:tan^{-1}(\frac{x+2}{\sqrt6})+C$

Explanation:

The denominator is irreducible, which means that we cannot use partial fractions to identify the coefficients of the separable fractions. The only method we can use is to complete the square and use the identity of inverse trigonometry.

Complete the square for the denominator. This is done by taking the square of half of the middle term, and then subtracting this digit on the end.

x^2+4x+10 = (x^2+4x+4)+10 -(4)

Factorize the parabolic function in the parentheses and simplify.

x^2+4x+10 = (x+2)^2+6

Rewrite the integral and pull out the six in front of the integral.

$\int \frac{6}{x^2+4x+10}\:dx = 6\int \frac{1}{(x+2)^2+6}\:dx$

Write the integral property of inverse tangent.

$\frac{1}{A} tan^{-1}(\frac{X}{A})+C = \int \frac{1}{X^2+A^2}$

By this rule, substitute the terms and simplify to get the and terms.

X^2 = (x+2)^2

X=x+2

A^2 = 6

$A=\sqrt6$

Substitute the and terms into the inverse tangent rule.

$\frac{1}{\sqrt6} tan^{-1}(\frac{x+2}{\sqrt6})+C$

Rationalize the denominator of the coefficient.

$\frac{1}{\sqrt6} \cdot \frac{\sqrt6}{\sqrt6} tan^{-1}(\frac{x+2}{\sqrt6})+C = \frac{\sqrt6}{6} tan^{-1}(\frac{x+2}{\sqrt6})+C$

Substituting this back in the original integral, this means that:

$6\int \frac{1}{(x+2)^2+6}\:dx = 6(\frac{\sqrt6}{6} tan^{-1}(\frac{x+2}{\sqrt6})+C)$

Do not forget to multiply the six that is outside the integral.

$= 6(\frac{\sqrt6}{6} tan^{-1}(\frac{x+2}{\sqrt6})+C) = \sqrt6\:tan^{-1}(\frac{x+2}{\sqrt6})+C$

The answer is: $\sqrt6\:tan^{-1}(\frac{x+2}{\sqrt6})+C$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Indefinite Integrals

Example Question #291 : Finding Integrals

Example Question #641 : Integrals

Example Question #17 : Indefinite Integrals

Example Question #11 : Indefinite Integrals

Example Question #11 : Indefinite Integrals

Example Question #12 : Indefinite Integrals

Example Question #21 : Indefinite Integrals

Example Question #21 : Indefinite Integrals

Example Question #23 : Indefinite Integrals

All Calculus 2 Resources

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