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High School Math : Integrals

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Example Questions

Example Question #1 : Finding Definite Integrals

Evaluate:

$\int_{1}^{e^{100}} \frac{1}{x}dx$

Possible Answers:

$\frac{1}{100}$

100

$\frac{100}{e}$

100e

$\ln 100$

Correct answer:

100

Explanation:

$\int_{1}^{e^{100}} \frac{dx}{x} = \ln e^{100} - \ln 1 = 100-0 = 100$

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Example Question #1 : Integrals

Find $\int_{0}^{\pi} (\cos^{2}x+\sin^{2}x)dx$

Possible Answers:

$2\pi$

$\pi$

Correct answer:

$\pi$

Explanation:

This is most easily solved by recognizing that $\sin^{2}x+\cos^{2}x=1$ .

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Example Question #1 : Integrals

$\int_{\pi}^{2\pi}\sin(2x){\mathrm{d} x}=?$

Possible Answers:

$\frac{1}{2}$

$\pi$

Correct answer:

Explanation:

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\sin(2x)$ , we can't use the power rule. Instead we end up with:

$\int f(x){\mathrm{d} x}=\int \sin(2x){\mathrm{d} x}=\frac{-\cos(2x)}{2}+c$

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\frac{-\cos(2b)}{2}+c)-(\frac{-\cos(2a)}{2}+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{\pi}^{2\pi}f(x){\mathrm{d} x}=(\frac{-\cos(4\pi)}{2})-(\frac{-\cos(2\pi)}{2})$

$\int_{\pi}^{2\pi}f(x){\mathrm{d} x}=(\frac{-1}{2})-(\frac{-1}{2})$

$\int_{\pi}^{2\pi}f(x){\mathrm{d} x}=0$

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Example Question #2 : Integrals

$\int_{2}^{4}x^2{\mathrm{d} x}=?$

Possible Answers:

$5\tfrac{1}{2}$

$22\tfrac{1}{3}$

$28\tfrac{2}{3}$

$11\tfrac{3}{4}$

$18\frac{2}{3}$

Correct answer:

$18\frac{2}{3}$

Explanation:

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our f(x)=x^2 , we can use the reverse power rule to find the indefinite integral or anti-derivative of our function:

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

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\frac{1}{3}b^3+c)-(\frac{1}{3}a^3+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{2}^{4}f(x){\mathrm{d} x}=(\frac{1}{3}\cdot 4^3)-(\frac{1}{3}\cdot 8^3)$

$\int_{2}^{4}f(x){\mathrm{d} x}=(\frac{64}{3})-(\frac{8}{3})$

$\int_{2}^{4}f(x){\mathrm{d} x}=\frac{56}{3}=18\frac{2}{3}$

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Example Question #43 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

$\int_{1}^{2}\frac{1}{x}{\mathrm{d} x}=?$

Possible Answers:

0.81

1.2

0.30

0.75

0.69

Correct answer:

0.69

Explanation:

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

As it turns out, since our $f(x)=\frac{1}{x}$ , the power rule really doesn't help us. $\frac{1}{x}$ has a special anti derivative: $\ln{x}$ .

$\int \frac{1}{x}{\mathrm{d} x}=\ln{x}+c$

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\ln{b}+c)-(\ln{a}+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{1}^{2}f(x){\mathrm{d} x}=(\ln(2))-(\ln(1))$

$\int_{1}^{2}f(x){\mathrm{d} x}=(\ln(2))-(0)$

$\int_{1}^{2}f(x){\mathrm{d} x}\approx 0.69$

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Example Question #44 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

$\int_{3}^{40}e^x{\mathrm{d} x}=?$

Possible Answers:

$2.35\cdot 10^{17}$

$1.05\cdot 10^{17}$

$2.35\cdot 10^{19}$

$2.35\cdot 10^{15}$

$2.35\cdot 10^{9}$

Correct answer:

$2.35\cdot 10^{17}$

Explanation:

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

As it turns out, since our f(x)=e^x , the power rule really doesn't help us. e^x is the only function that is it's OWN anti-derivative. That means we're still going to be working with e^x .

$\int e^x{\mathrm{d} x}=e^x+c$

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(e^b+c)-(e^a+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{3}^{40}f(x){\mathrm{d} x}=(e^{40})-(e^3)$

$\int_{3}^{40}f(x){\mathrm{d} x}=(2.35\cdot 10^{17})-(20.09)$

Because e^3 is so small in comparison to the value we got for $e^{40}$ , our answer will end up being $2.35\cdot 10^{17}$

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Example Question #31 : Calculus Ii — Integrals

$\int_2^6(5x^2+3x+2){\mathrm{d} x}=?$

Possible Answers:

$401\tfrac{1}{3}$

$26\tfrac{2}{3}$

$204\tfrac{2}{3}$

$402\tfrac{2}{3}$

$-12\tfrac{2}{3}$

Correct answer:

$402\tfrac{2}{3}$

Explanation:

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our f(x)=5x^2+3x+2 , we can use the power rule for all of the terms involved to find our anti-derivative:

$\int (5x^2+3x+2){\mathrm{d} x}=\frac{5}{3}x^3+\frac{3}{2}x^2+2x+c$

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\frac{5}{3}b^3+\frac{3}{2}b^2+2b+c)-(\frac{5}{3}a^3+\frac{3}{2}a^2+2a+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{2}^{6}f(x){\mathrm{d} x}=(\frac{5}{3}(6)^3+\frac{3}{2}(6)^2+2(6))-(\frac{5}{3}(2)^3+\frac{3}{2}(2)^2+2(2))$

$\int_{2}^{6}f(x){\mathrm{d} x}=(360+54+12)-(\frac{40}{3}+6+4)$

$\int_{2}^{6}f(x){\mathrm{d} x}=(426)-(23\tfrac{1}{3})$

$\int_{2}^{6}f(x){\mathrm{d} x}=402\tfrac{2}{3}$

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Example Question #2 : Finding Definite Integrals

$\int_3^5\frac{x+3}{x}{\mathrm{d} x}=?$

Possible Answers:

3.21

5.67

1.33

6.12

3.53

Correct answer:

3.53

Explanation:

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\frac{x+3}{x}$ , we can't use the power rule. We have to break up the quotient into separate parts:

$f(x)=\frac{x}{x}+\frac{3}{x}=1+\frac{3}{x}$ .

The integral of 1 should be no problem, but the other half is a bit more tricky:

$\int\frac{3}{x}{\mathrm{d} x}$ is really the same as $3\int\frac{1}{x}{\mathrm{d} x}$ . Since $\int\frac{1}{x}{\mathrm{d} x}=\ln{x}$ , $\int\frac{3}{x}{\mathrm{d} x}=3\ln{x}$ .

Therefore:

$\int\frac{x+3}{x}{\mathrm{d} x}=x+3\ln{x}+c$

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(b+3\ln{b}+c)-(a+3\ln{a}+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{3}^{5}f(x){\mathrm{d} x}=(5+3\ln{5})-(3+3\ln{3})$

$\int_{3}^{5}f(x){\mathrm{d} x}=(9.83)-(6.30)$

$\int_{3}^{5}f(x){\mathrm{d} x}\approx 3.53$

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Example Question #5 : Integrals

$\int_2^5\sqrt{x}\text{ }{\mathrm{d} x}=?$

Possible Answers:

2.59

5.56

1.89

9.34

7.52

Correct answer:

5.56

Explanation:

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\sqrt{x}$ , we can use the power rule, if we turn it into an exponent:

$f(x)=\sqrt{x}=x^{\frac{1}{2}}$

This means that:

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

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(\frac{2}{3}b^\frac{3}{2}+c)-(\frac{2}{3}a^\frac{3}{2}+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{2}^{5}f(x){\mathrm{d} x}=(\frac{2}{3}(5)^\frac{3}{2})-(\frac{2}{3}(2)^\frac{3}{2})$

$\int_{2}^{5}f(x){\mathrm{d} x}\approx(7.45)-(1.89)$

$\int_{2}^{5}f(x){\mathrm{d} x}\approx5.56$

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Example Question #6 : Integrals

$\int_{12}^{15}\ln(x){\mathrm{d} x}=?$

Possible Answers:

8.7

71.12

3.9

10.8

7.8

Correct answer:

7.8

Explanation:

Remember the fundamental theorem of calculus!

$\int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $f(x)=\ln(x)$ , we can't use the power rule, as it has a special antiderivative:

$\int f(x){\mathrm{d} x}=\int \ln(x){\mathrm{d} x}=x\ln(x)-x+c$

Remember to include the + c for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\int_{a}^{b}f(x){\mathrm{d} x}=(b\ln(b)-b+c)-(a\ln(a)-a+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\int_{12}^{15}f(x){\mathrm{d} x}=(15\ln(15)-15)-(12\ln(12)-12)$

$\int_{12}^{15}f(x){\mathrm{d} x}\approx(25.62)-(17.82)$

$\int_{12}^{15}f(x){\mathrm{d} x}\approx7.8$

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High School Math : Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Finding Definite Integrals

Example Question #1 : Integrals

Example Question #1 : Integrals

Example Question #2 : Integrals

Example Question #43 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #44 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #31 : Calculus Ii — Integrals

Example Question #2 : Finding Definite Integrals

Example Question #5 : Integrals

Example Question #6 : Integrals

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