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Calculus 2 : Integrals

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Example Questions

Example Question #342 : Finding Integrals

Evaluate.

$\int -\cos(7x) \ dx$ \(\displaystyle \int -\cos(7x) \ dx\)

Possible Answers:

$F(x) = - \sin(7x)$ \(\displaystyle F(x) = - \sin(7x)\)

$F(x) = \frac{\sin(x)}{7}$ \(\displaystyle F(x) = \frac{\sin(x)}{7}\)

$F(x) = - \frac{\sin(7x)}{7}$ \(\displaystyle F(x) = - \frac{\sin(7x)}{7}\)

Answer not listed.

$F(x) = \frac{\cos(7x)}{7}$ \(\displaystyle F(x) = \frac{\cos(7x)}{7}\)

Correct answer:

$F(x) = - \frac{\sin(7x)}{7}$ \(\displaystyle F(x) = - \frac{\sin(7x)}{7}\)

Explanation:

$\int f(x) \ dx$ \(\displaystyle \int f(x) \ dx\)

In order to evaluate this integral, first find the antiderivative of f(x). \(\displaystyle f(x).\)

If \(\displaystyle f(x) = a\) then \(\displaystyle F(x) = ax + C\)

If \(\displaystyle f(x) = x a\) then $F(x) = \frac{x^{a+1}}{a+1} + C$ \(\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C\)

If $f(x) = \frac{1}{x}$ \(\displaystyle f(x) = \frac{1}{x}\) then \(\displaystyle F(x) = ln(x) + C\)

If \(\displaystyle f(x) = e ^x\) then \(\displaystyle F(x) = e^x + C\)

If $f(x) = \cos(x)$ \(\displaystyle f(x) = \cos(x)\) then $F(x) = \sin(x) + C$ \(\displaystyle F(x) = \sin(x) + C\)

If $f(x) = \sin(x)$ \(\displaystyle f(x) = \sin(x)\) then $F(x) = - \cos(x) + C$ \(\displaystyle F(x) = - \cos(x) + C\)

If $f(x) = \sec^2 (x)$ \(\displaystyle f(x) = \sec^2 (x)\) then $F(x) = \tan(x) + C$ \(\displaystyle F(x) = \tan(x) + C\)

In this case, $f(x) =-\cos(7x)$ \(\displaystyle f(x) =-\cos(7x)\).

The antiderivative is $F(x) = - \frac{\sin(7x)}{7}$ \(\displaystyle F(x) = - \frac{\sin(7x)}{7}\).

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Example Question #343 : Finding Integrals

Evaluate.

$\int e^{2x-5} - \sin(2x-5) \ dx$ \(\displaystyle \int e^{2x-5} - \sin(2x-5) \ dx\)

Possible Answers:

$F(x) = \frac{e^{2x-5}}{x} - \frac{\cos(2x-5)}{x}$ \(\displaystyle F(x) = \frac{e^{2x-5}}{x} - \frac{\cos(2x-5)}{x}\)

$F(x) = e^{2x-5}+ \cos(2x-5)$ \(\displaystyle F(x) = e^{2x-5}+ \cos(2x-5)\)

$F(x) = \frac{e^{2x-5}}{2} + \frac{\cos(2x-5)}{2}$ \(\displaystyle F(x) = \frac{e^{2x-5}}{2} + \frac{\cos(2x-5)}{2}\)

$F(x) = \frac{e^{2x-5}}{e^2} + \frac{\cos(2x-5)}{\cos2}$ \(\displaystyle F(x) = \frac{e^{2x-5}}{e^2} + \frac{\cos(2x-5)}{\cos2}\)

Answer not listed.

Correct answer:

$F(x) = \frac{e^{2x-5}}{2} + \frac{\cos(2x-5)}{2}$ \(\displaystyle F(x) = \frac{e^{2x-5}}{2} + \frac{\cos(2x-5)}{2}\)

Explanation:

$\int f(x) \ dx$ \(\displaystyle \int f(x) \ dx\)

In order to evaluate this integral, first find the antiderivative of f(x). \(\displaystyle f(x).\)

If \(\displaystyle f(x) = a\) then \(\displaystyle F(x) = ax + C\)

If \(\displaystyle f(x) = x a\) then $F(x) = \frac{x^{a+1}}{a+1} + C$ \(\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C\)

If $f(x) = \frac{1}{x}$ \(\displaystyle f(x) = \frac{1}{x}\) then \(\displaystyle F(x) = ln(x) + C\)

If \(\displaystyle f(x) = e ^x\) then \(\displaystyle F(x) = e^x + C\)

If $f(x) = \cos(x)$ \(\displaystyle f(x) = \cos(x)\) then $F(x) = \sin(x) + C$ \(\displaystyle F(x) = \sin(x) + C\)

If $f(x) = \sin(x)$ \(\displaystyle f(x) = \sin(x)\) then $F(x) = - \cos(x) + C$ \(\displaystyle F(x) = - \cos(x) + C\)

If $f(x) = \sec^2 (x)$ \(\displaystyle f(x) = \sec^2 (x)\) then $F(x) = \tan(x) + C$ \(\displaystyle F(x) = \tan(x) + C\)

In this case, $f(x) =e^{2x-5} - \sin(2x-5)$ \(\displaystyle f(x) =e^{2x-5} - \sin(2x-5)\).

The antiderivative is $F(x) = \frac{e^{2x-5}}{2} + \frac{\cos(2x-5)}{2}$ \(\displaystyle F(x) = \frac{e^{2x-5}}{2} + \frac{\cos(2x-5)}{2}\).

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Example Question #344 : Finding Integrals

Evaluate.

$\int 5x^4 + 12x \ dx$ \(\displaystyle \int 5x^4 + 12x \ dx\)

Possible Answers:

\(\displaystyle F(x) = 5x^5 + 12x^2\)

$F(x) = \frac{1}{5}x^5 + \frac{1}{12}x^2$ \(\displaystyle F(x) = \frac{1}{5}x^5 + \frac{1}{12}x^2\)

\(\displaystyle F(x) = x^5 + 6x^2\)

\(\displaystyle F(x) = 5x^4 + 12x\)

Answer not listed.

Correct answer:

\(\displaystyle F(x) = x^5 + 6x^2\)

Explanation:

$\int f(x) \ dx$ \(\displaystyle \int f(x) \ dx\)

In order to evaluate this integral, first find the antiderivative of f(x). \(\displaystyle f(x).\)

If \(\displaystyle f(x) = a\) then \(\displaystyle F(x) = ax + C\)

If \(\displaystyle f(x) = x a\) then $F(x) = \frac{x^{a+1}}{a+1} + C$ \(\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C\)

If $f(x) = \frac{1}{x}$ \(\displaystyle f(x) = \frac{1}{x}\) then \(\displaystyle F(x) = ln(x) + C\)

If \(\displaystyle f(x) = e ^x\) then \(\displaystyle F(x) = e^x + C\)

If $f(x) = \cos(x)$ \(\displaystyle f(x) = \cos(x)\) then $F(x) = \sin(x) + C$ \(\displaystyle F(x) = \sin(x) + C\)

If $f(x) = \sin(x)$ \(\displaystyle f(x) = \sin(x)\) then $F(x) = - \cos(x) + C$ \(\displaystyle F(x) = - \cos(x) + C\)

If $f(x) = \sec^2 (x)$ \(\displaystyle f(x) = \sec^2 (x)\) then $F(x) = \tan(x) + C$ \(\displaystyle F(x) = \tan(x) + C\)

In this case, \(\displaystyle f(x) =5x^4 + 12x\).

The antiderivative is \(\displaystyle F(x) = x^5 + 6x^2\).

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Example Question #345 : Finding Integrals

Evaluate.

$\int \frac{4}{x-1} \ dx$ \(\displaystyle \int \frac{4}{x-1} \ dx\)

Possible Answers:

$F(x) =\frac{\ln(x-1)}{4}$ \(\displaystyle F(x) =\frac{\ln(x-1)}{4}\)

$F(x) = \ln(x-1)$ \(\displaystyle F(x) = \ln(x-1)\)

Answer not listed.

$F(x) = 4 \ln(x-1)$ \(\displaystyle F(x) = 4 \ln(x-1)\)

$F(x) = \frac{1}{ \ln(x-1)}$ \(\displaystyle F(x) = \frac{1}{ \ln(x-1)}\)

Correct answer:

$F(x) = 4 \ln(x-1)$ \(\displaystyle F(x) = 4 \ln(x-1)\)

Explanation:

$\int f(x) \ dx$ \(\displaystyle \int f(x) \ dx\)

In order to evaluate this integral, first find the antiderivative of f(x). \(\displaystyle f(x).\)

If \(\displaystyle f(x) = a\) then \(\displaystyle F(x) = ax + C\)

If \(\displaystyle f(x) = x a\) then $F(x) = \frac{x^{a+1}}{a+1} + C$ \(\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C\)

If $f(x) = \frac{1}{x}$ \(\displaystyle f(x) = \frac{1}{x}\) then \(\displaystyle F(x) = ln(x) + C\)

If \(\displaystyle f(x) = e ^x\) then \(\displaystyle F(x) = e^x + C\)

If $f(x) = \cos(x)$ \(\displaystyle f(x) = \cos(x)\) then $F(x) = \sin(x) + C$ \(\displaystyle F(x) = \sin(x) + C\)

If $f(x) = \sin(x)$ \(\displaystyle f(x) = \sin(x)\) then $F(x) = - \cos(x) + C$ \(\displaystyle F(x) = - \cos(x) + C\)

If $f(x) = \sec^2 (x)$ \(\displaystyle f(x) = \sec^2 (x)\) then $F(x) = \tan(x) + C$ \(\displaystyle F(x) = \tan(x) + C\)

In this case, $f(x) = \frac{4}{x-1}$ \(\displaystyle f(x) = \frac{4}{x-1}\).

The antiderivative is $F(x) = 4 \ln(x-1)$ \(\displaystyle F(x) = 4 \ln(x-1)\).

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Example Question #346 : Finding Integrals

Evaluate.

$\int 12 \ dx$ \(\displaystyle \int 12 \ dx\)

Possible Answers:

\(\displaystyle F(x) = 12x\)

\(\displaystyle F(x) = 12x^2\)

F(x) =0 \(\displaystyle F(x) =0\)

Answer not listed

$F(x) = \frac{1}{12}x$ \(\displaystyle F(x) = \frac{1}{12}x\)

Correct answer:

\(\displaystyle F(x) = 12x\)

Explanation:

$\int f(x) \ dx$ \(\displaystyle \int f(x) \ dx\)

In order to evaluate this integral, first find the antiderivative of f(x). \(\displaystyle f(x).\)

If \(\displaystyle f(x) = a\) then \(\displaystyle F(x) = ax + C\)

If \(\displaystyle f(x) = x a\) then $F(x) = \frac{x^{a+1}}{a+1} + C$ \(\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C\)

If $f(x) = \frac{1}{x}$ \(\displaystyle f(x) = \frac{1}{x}\) then \(\displaystyle F(x) = ln(x) + C\)

If \(\displaystyle f(x) = e ^x\) then \(\displaystyle F(x) = e^x + C\)

If $f(x) = \cos(x)$ \(\displaystyle f(x) = \cos(x)\) then $F(x) = \sin(x) + C$ \(\displaystyle F(x) = \sin(x) + C\)

If $f(x) = \sin(x)$ \(\displaystyle f(x) = \sin(x)\) then $F(x) = - \cos(x) + C$ \(\displaystyle F(x) = - \cos(x) + C\)

If $f(x) = \sec^2 (x)$ \(\displaystyle f(x) = \sec^2 (x)\) then $F(x) = \tan(x) + C$ \(\displaystyle F(x) = \tan(x) + C\)

In this case, \(\displaystyle f(x) = 12\).

The antiderivative is \(\displaystyle F(x) = 12x\).

Report an Error

Example Question #347 : Finding Integrals

Evaluate.

$\int -\sin(6x) +3 \ dx$ \(\displaystyle \int -\sin(6x) +3 \ dx\)

Possible Answers:

Answer not listed.

$F(x) = \frac{\cos(6x)}{6} + 3x$ \(\displaystyle F(x) = \frac{\cos(6x)}{6} + 3x\)

$F(x) = \frac{\cos(x)}{6x} + 3$ \(\displaystyle F(x) = \frac{\cos(x)}{6x} + 3\)

$F(x) = -\cos(6x) + 3x$ \(\displaystyle F(x) = -\cos(6x) + 3x\)

$F(x) =- \frac{\cos(6x)}{6} + 3x$ \(\displaystyle F(x) =- \frac{\cos(6x)}{6} + 3x\)

Correct answer:

$F(x) = \frac{\cos(6x)}{6} + 3x$ \(\displaystyle F(x) = \frac{\cos(6x)}{6} + 3x\)

Explanation:

$\int f(x) \ dx$ \(\displaystyle \int f(x) \ dx\)

In order to evaluate this integral, first find the antiderivative of f(x). \(\displaystyle f(x).\)

If \(\displaystyle f(x) = a\) then \(\displaystyle F(x) = ax + C\)

If \(\displaystyle f(x) = x a\) then $F(x) = \frac{x^{a+1}}{a+1} + C$ \(\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C\)

If $f(x) = \frac{1}{x}$ \(\displaystyle f(x) = \frac{1}{x}\) then \(\displaystyle F(x) = ln(x) + C\)

If \(\displaystyle f(x) = e ^x\) then \(\displaystyle F(x) = e^x + C\)

If $f(x) = \cos(x)$ \(\displaystyle f(x) = \cos(x)\) then $F(x) = \sin(x) + C$ \(\displaystyle F(x) = \sin(x) + C\)

If $f(x) = \sin(x)$ \(\displaystyle f(x) = \sin(x)\) then $F(x) = - \cos(x) + C$ \(\displaystyle F(x) = - \cos(x) + C\)

If $f(x) = \sec^2 (x)$ \(\displaystyle f(x) = \sec^2 (x)\) then $F(x) = \tan(x) + C$ \(\displaystyle F(x) = \tan(x) + C\)

In this case, $f(x) = -\sin(6x) +3$ \(\displaystyle f(x) = -\sin(6x) +3\).

The antiderivative is $F(x) = \frac{\cos(6x)}{6} + 3x$ \(\displaystyle F(x) = \frac{\cos(6x)}{6} + 3x\).

Report an Error

Example Question #348 : Finding Integrals

Evaluate.

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

Possible Answers:

$F(x) = x^2 - \frac{\ln(x)}{3}$ \(\displaystyle F(x) = x^2 - \frac{\ln(x)}{3}\)

Answer not listed.

$F(x) = 2x^2 - \frac{\ln(3x)}{3}$ \(\displaystyle F(x) = 2x^2 - \frac{\ln(3x)}{3}\)

$F(x) = 2x^2 - \ln(x)$ \(\displaystyle F(x) = 2x^2 - \ln(x)\)

$F(x) = \frac{1}{2}x^2 - \ln(x)$ \(\displaystyle F(x) = \frac{1}{2}x^2 - \ln(x)\)

Correct answer:

$F(x) = x^2 - \frac{\ln(x)}{3}$ \(\displaystyle F(x) = x^2 - \frac{\ln(x)}{3}\)

Explanation:

$\int f(x) \ dx$ \(\displaystyle \int f(x) \ dx\)

In order to evaluate this integral, first find the antiderivative of f(x). \(\displaystyle f(x).\)

If \(\displaystyle f(x) = a\) then \(\displaystyle F(x) = ax + C\)

If \(\displaystyle f(x) = x a\) then $F(x) = \frac{x^{a+1}}{a+1} + C$ \(\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C\)

If $f(x) = \frac{1}{x}$ \(\displaystyle f(x) = \frac{1}{x}\) then \(\displaystyle F(x) = ln(x) + C\)

If \(\displaystyle f(x) = e ^x\) then \(\displaystyle F(x) = e^x + C\)

If $f(x) = \cos(x)$ \(\displaystyle f(x) = \cos(x)\) then $F(x) = \sin(x) + C$ \(\displaystyle F(x) = \sin(x) + C\)

If $f(x) = \sin(x)$ \(\displaystyle f(x) = \sin(x)\) then $F(x) = - \cos(x) + C$ \(\displaystyle F(x) = - \cos(x) + C\)

If $f(x) = \sec^2 (x)$ \(\displaystyle f(x) = \sec^2 (x)\) then $F(x) = \tan(x) + C$ \(\displaystyle F(x) = \tan(x) + C\)

In this case, $f(x) = 2x - \frac{1}{3x}$ \(\displaystyle f(x) = 2x - \frac{1}{3x}\).

The antiderivative is $F(x) = x^2 - \frac{\ln(x)}{3}$ \(\displaystyle F(x) = x^2 - \frac{\ln(x)}{3}\).

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Example Question #691 : Integrals

Evaluate.

$\int e^{7x} - 7 \ dx$ \(\displaystyle \int e^{7x} - 7 \ dx\)

Possible Answers:

Answer not listed.

$F(x) = 7e^{7x}$ \(\displaystyle F(x) = 7e^{7x}\)

$F(x) = 7e^{7x} - 7$ \(\displaystyle F(x) = 7e^{7x} - 7\)

$F(x) = \frac{e^{7x}}{7} - 7x$ \(\displaystyle F(x) = \frac{e^{7x}}{7} - 7x\)

$F(x) = e^{7x} - 7$ \(\displaystyle F(x) = e^{7x} - 7\)

Correct answer:

$F(x) = \frac{e^{7x}}{7} - 7x$ \(\displaystyle F(x) = \frac{e^{7x}}{7} - 7x\)

Explanation:

$\int f(x) \ dx$ \(\displaystyle \int f(x) \ dx\)

In order to evaluate this integral, first find the antiderivative of f(x). \(\displaystyle f(x).\)

If \(\displaystyle f(x) = a\) then \(\displaystyle F(x) = ax + C\)

If \(\displaystyle f(x) = x a\) then $F(x) = \frac{x^{a+1}}{a+1} + C$ \(\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C\)

If $f(x) = \frac{1}{x}$ \(\displaystyle f(x) = \frac{1}{x}\) then \(\displaystyle F(x) = ln(x) + C\)

If \(\displaystyle f(x) = e ^x\) then \(\displaystyle F(x) = e^x + C\)

If $f(x) = \cos(x)$ \(\displaystyle f(x) = \cos(x)\) then $F(x) = \sin(x) + C$ \(\displaystyle F(x) = \sin(x) + C\)

If $f(x) = \sin(x)$ \(\displaystyle f(x) = \sin(x)\) then $F(x) = - \cos(x) + C$ \(\displaystyle F(x) = - \cos(x) + C\)

If $f(x) = \sec^2 (x)$ \(\displaystyle f(x) = \sec^2 (x)\) then $F(x) = \tan(x) + C$ \(\displaystyle F(x) = \tan(x) + C\)

In this case, $f(x) = e^{7x} + 7$ \(\displaystyle f(x) = e^{7x} + 7\).

The antiderivative is $F(x) = \frac{e^{7x}}{7} - 7x$ \(\displaystyle F(x) = \frac{e^{7x}}{7} - 7x\).

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Example Question #72 : Indefinite Integrals

Evaluate.

$\int \sin(x) + 6 - \frac{1}{5x}\ dx$ \(\displaystyle \int \sin(x) + 6 - \frac{1}{5x}\ dx\)

Possible Answers:

$F(x) = - \cos(x) + 6x - 5\ln (x)$ \(\displaystyle F(x) = - \cos(x) + 6x - 5\ln (x)\)

$F(x) = - \cos(x) + 6x - \frac{\ln (x)}{5}$ \(\displaystyle F(x) = - \cos(x) + 6x - \frac{\ln (x)}{5}\)

Answer not listed.

$F(x) = - \cos(x) + 6 - \frac{\ln (5x)}{5}$ \(\displaystyle F(x) = - \cos(x) + 6 - \frac{\ln (5x)}{5}\)

$F(x) = - \sin(x) + 6x - \frac{\ln (5x)}{5}$ \(\displaystyle F(x) = - \sin(x) + 6x - \frac{\ln (5x)}{5}\)

Correct answer:

$F(x) = - \cos(x) + 6x - \frac{\ln (x)}{5}$ \(\displaystyle F(x) = - \cos(x) + 6x - \frac{\ln (x)}{5}\)

Explanation:

\(\displaystyle f(x) = a\) then \(\displaystyle F(x) = ax + C\)

If \(\displaystyle f(x) = x a\) then $F(x) = \frac{x^{a+1}}{a+1} + C$ \(\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C\)

If $f(x) = \frac{1}{x}$ \(\displaystyle f(x) = \frac{1}{x}\) then \(\displaystyle F(x) = ln(x) + C\)

If \(\displaystyle f(x) = e ^x\) then \(\displaystyle F(x) = e^x + C\)

If $f(x) = \cos(x)$ \(\displaystyle f(x) = \cos(x)\) then $F(x) = \sin(x) + C$ \(\displaystyle F(x) = \sin(x) + C\)

If $f(x) = \sin(x)$ \(\displaystyle f(x) = \sin(x)\) then $F(x) = - \cos(x) + C$ \(\displaystyle F(x) = - \cos(x) + C\)

If $f(x) = \sec^2 (x)$ \(\displaystyle f(x) = \sec^2 (x)\) then $F(x) = \tan(x) + C$ \(\displaystyle F(x) = \tan(x) + C\)

In this case, $f(x) = \sin(x) + 6 - \frac{1}{5x}$ \(\displaystyle f(x) = \sin(x) + 6 - \frac{1}{5x}\).

The antiderivative is $F(x) = - \cos(x) + 6x - \frac{\ln (x)}{5}$ \(\displaystyle F(x) = - \cos(x) + 6x - \frac{\ln (x)}{5}\).

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Example Question #692 : Integrals

$\int \frac{2x^2-8x+2}{2}dx$ \(\displaystyle \int \frac{2x^2-8x+2}{2}dx\)

Possible Answers:

$\frac{x^3}{3}-2x^2+x$ \(\displaystyle \frac{x^3}{3}-2x^2+x\)

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

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

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

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

Correct answer:

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

Explanation:

First, make the fraction three separate terms: $\int x^2-4x+1dx$ \(\displaystyle \int x^2-4x+1dx\). Now, integrate as normal, remembering to raise the exponent by 1 and then also putting that result on the denominator: $\frac{x^3}{3}-2x^2+x$ \(\displaystyle \frac{x^3}{3}-2x^2+x\). Remember to add a C at the end because it is an indefinite integral: $\frac{x^3}{3}-2x^2+x+C$ \(\displaystyle \frac{x^3}{3}-2x^2+x+C\).

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