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Calculus 2 : Definite Integrals

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Example Question #11 : Definite Integrals

Find the integral of this equation:

$\int_{0}^{2} (x+2)(x^2+4)dx$

Possible Answers:

$\frac{97}{3}$

$\frac{100}{3}$

$\frac{103}{3}$

$\frac{450}{12}$

$\frac{425}{12}$

Correct answer:

$\frac{100}{3}$

Explanation:

Distribute the polynomial first:
(x+2)(x^2+4) = x^3 + 2x^2 + 4x + 8

Then integrate:

$\int_{0}^{2}x^3 + 2x^2 + 4x + 8 \ dx = \Big[\frac{x^4}{4} + \frac{2x^3}{3}+2x^2+8x\Big]_0 ^2$

And plug and solve:

$\frac{16}{4} + \frac{16}{3} + 8 + 16 - 0=\frac{100}{3}$

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Example Question #12 : Definite Integrals

Evaluate the integral:

$\int_{0}^{3} (x+1)(x-2)dx$

Possible Answers:

$-\frac{3}{2}$

$\frac{3}{2}$

Correct answer:

$-\frac{3}{2}$

Explanation:

Factor the polynomial:

(x+1)(x-2) = x^2 - x -2

Then integrate (To integrate, add 1 power of each term. So $x^2\rightarrow x^3$ . Then divide the term by the value of the exponent. $x^3 \rightarrow \frac{1}{3}x^3$ .

$\int_{0}^{3} (x^2 - x -3 ) = \Big[\frac{1}{3}x^3-\frac{1}{2}x^2-2x\Big]_{0}^{3}$

Then evaulate the integral:

$[\frac{1}{3}(3)^3 - \frac{1}{2}(3)^3-2(3)] - [\frac{1}{3}(0)^3 - \frac{1}{2}(0)^3-2(0)]$

$= 9 - \frac{9}{2} - 6$

= $-\frac{3}{2}$

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Example Question #13 : Definite Integrals

$\int_{0}^{5} \frac{x+5}{x^2+x+-20} dx$

Possible Answers:

-ln(4)

ln(-4)

ln(2)

-ln(2)

ln(4)

Correct answer:

-ln(4)

Explanation:

Simplify:

$\frac{x+5}{x^2+x-20} = \frac{x+5}{(x+5)(x-4)} = \frac{1}{x-4}$

Integrate:

$\int_{0}^{5}\frac{1}{x-4}dx = \Big[ln(x-4)\Big]_{0}^{5}$

Evaluate:

= ln|(5) - 4| - ln|(0) - 4|

= -ln(4)

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Example Question #14 : Definite Integrals

Find the integral:

$\int_{0}^{5} \frac{x+2}{x^2+4x+4} dx$

Possible Answers:

$ln(\frac{2}{5})$

$ln(\frac{7}{2})$

$ln(\frac{7}{3})$

Correct answer:

$ln(\frac{7}{2})$

Explanation:

$\frac{x+2}{x^2+4x+4} = \frac{x+2}{(x+2)(x+2)} = \frac{1}{x+2}$

Integrate new equation:

$\int_{0}^{5} \frac{1}{x+2} dx = \Big[ln\left |(x+2) \right |\Big]_{0}^{5}$

$= ln\left |(5)+2\right | - ln\left |((0)+2)\right | = ln(7) - ln(2)$

Use natural log rules:

$ln(7)-ln(2) = ln(\frac{7}{2})$

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Example Question #15 : Definite Integrals

$\int_{1}^{3} \frac{x^2 +8x + 15}{x^2+3x} dx$

Evaulate the integral:

$\int_{1}^{3} \frac{x^2 +8x + 15}{x^2+3x} dx$

Possible Answers:

= 2+ 5ln(3)

= 2+ 5ln(2)

= 2 - 5ln(3)

= 4+ 5ln(3)

Correct answer:

= 2+ 5ln(3)

Explanation:

Simplify:

$\frac{x^2 +8x + 15}{x^2+3x} = \frac{(x+3)(x+5)}{x(x+3)} = \frac{x+5}{x}$

Seperate the terms:

$\frac{x+5}{x} = \frac{x}{x}+\frac{5}{x} = 1 + \frac{5}{x}$

Now integrate the simplified form:

$\int_{1}^{3} (1 + \frac{5}{x}) dx$

$= \Big[x + 5ln|x|\Big]_{1}^{3}$

= [3 + 5ln|3|] - [1 + 5ln|1|]

= 2+ 5ln(3)

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Example Question #16 : Definite Integrals

Evaluate the definite integral

$\int_{0}^{\pi /2} [sin(x)+cos(x)]\,dx$

Possible Answers:

$\frac{\pi}{2}$

$\pi/4$

Correct answer:

Explanation:

Because integration is a linear operation, we are able to anti-differentiate the function term by term.

We use the properties that

The anti-derivative of is
The anti-derivative of is

to solve the definite integral

$\int_{0}^{\pi /2} [sin(x)+cos(x)]\,dx=(-cos(x)+sin(x)\Big|_{0}^{\frac{\pi}{2}}$

And by the corollary of the Fundamental Theorem of Calculus

$(-cos(x)+sin(x)\Big|_{0}^{\frac{\pi}{2}}=[-cos(\frac{\pi}{2})+sin(\frac{\pi}{2})]-[-cos(0)+sin(0)]$

=[0+1]-[-1+0]=1+1=2

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Example Question #17 : Definite Integrals

Evaluate the definite integral

$\int_{1}^{4} \frac{1}{x}\,dx$

Possible Answers:

$-\frac{3}{4}$

$ln \, 4$

DNE

Correct answer:

$ln \, 4$

Explanation:

We use the property that

The antii-derivative of $\frac{1}{x}$ is

to solve the definite integral

$\int_{1}^{4} \frac{1}{x}\,dx=(ln|x|)\Big|_{1}^{4}$

And by the corollary of the Fundamental Theorem of Calculus

$(ln|x|)\Big|_{1}^{4}=[ln|4|]-[ln|1|]$

And since $ln\,1=0$

the definite integral becomes

$=ln \,4$

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Example Question #18 : Definite Integrals

Evaluate the following definite integral:

$\int_0^5 \frac{x}{x+1} dx$

Possible Answers:

$\frac{5}{6}$

$\frac{5}{6}-\ln(5)$

$5-\ln(6)$

$\ln(6)$

Correct answer:

$5-\ln(6)$

Explanation:

$\int_0^5 \frac{x}{x+1}dx$ is an integral of a rational function, so we need to make the denominator into one of the basic rational functions we know the antiderivatives of and then split it up into multiple parts if necessary.

The most obvious thing to do then, is to substitute y = x + 1 :

$y = x+1\;\;\;dy = dx$

$x = 0\rightarrow y=1\;\;\;x=5\rightarrow y=6$

So our new integral is:

$\int_1^6\frac{y-1}{y} dy$

which can then be split up into two pieces:

$\int_1^6 1 dy \;- \int_1^6 \frac{1}{y}dy$

both of these are simple:

$y\Big|^6_{y=1} - \ln y\Big|^6_{y=1}$

$(6-1)-(\ln6 -\ln1) = 5-\ln6$

And that's the final answer.

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Example Question #19 : Definite Integrals

Evaluate $\int^{1}_{-1}\tan(x)dx$ .

Possible Answers:

None of the other answers

$\frac{\pi^2}{2}$

$\pi$

$2\pi$

Correct answer:

None of the other answers

Explanation:

The correct answer is .

Instead of evaluating the integral directly using antiderivatives, it is much eaiser to notice that the bounds of integration are negatives of each other, and that $\tan(x)$ is an odd function

(in other words, $\tan(-x) = -\tan(x), -\frac{\pi}{2} \le x \le \frac{\pi}{2}$ )

Hence

$\int^{1}_{-1}\tan(x)dx = \int^{1}_{0}\tan(x)dx + \int^{0}_{-1}\tan(x)dx$

$= \int^{1}_{0}\tan(x)dx - \int^{1}_{0}\tan(x)dx$ . Since $\tan(x)$ is odd.

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Example Question #20 : Definite Integrals

Evaluate $\int^{e^2}_{e} \ln(x)dx$

Possible Answers:

e^2-e

None of the other answers

$\frac{e^2}{2}$

e^2

Correct answer:

e^2

Explanation:

We use integration by parts for this integral.

Let $u = \ln(x)$ , and dv = dx , then taking the derivative of and the antiderivative of , we get $du = \frac{1}{x}dx$ , and v=x .

Using the integration by parts formula, $uv - \int v du$ , we have

$\int^{e^2}_{e} \ln(x)dx$ . Start

$=[x\ln(x)]^{e^2}_{e}-\int^{e^2}_{e} dx$ . Use the integration formula. (do not evaluate the left part yet).

$=[x\ln(x)-x]^{e^2}_{e}$ . Evaluate the last integral, and begin to evaluate the entire expression for the antiderivative.

$=(e^2\ln(e^2)-e^2) - (e\ln(e)-e)$ . Evaluate

=(2e^2-e^2)-(e-e) . Use rules of logs

=e^2 . Simplify

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Calculus 2 : Definite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Definite Integrals

Example Question #12 : Definite Integrals

Example Question #13 : Definite Integrals

Example Question #14 : Definite Integrals

Example Question #15 : Definite Integrals

Example Question #16 : Definite Integrals

Example Question #17 : Definite Integrals

Example Question #18 : Definite Integrals

Example Question #19 : Definite Integrals

Example Question #20 : Definite Integrals

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