High School Math : Finding Definite Integrals

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #1 : Finding Definite Integrals

Evaluate:

\(\displaystyle \int_{1}^{e^{100}} \frac{1}{x}dx\)

Possible Answers:

\(\displaystyle \frac{100}{e}\)

\(\displaystyle \frac{1}{100}\)

\(\displaystyle 100\)

\(\displaystyle \ln 100\)

\(\displaystyle 100e\)

Correct answer:

\(\displaystyle 100\)

Explanation:

\(\displaystyle \int_{1}^{e^{100}} \frac{dx}{x} = \ln e^{100} - \ln 1 = 100-0 = 100\) 

Example Question #2 : Finding Definite Integrals

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

Possible Answers:

\(\displaystyle 3\)

\(\displaystyle \pi\)

\(\displaystyle 1\)

\(\displaystyle 2\pi\)

Correct answer:

\(\displaystyle \pi\)

Explanation:

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

Example Question #3 : Finding Definite Integrals

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

Possible Answers:

\(\displaystyle \frac{1}{2}\)

\(\displaystyle 2\)

\(\displaystyle \pi\)

\(\displaystyle 0\)

\(\displaystyle 1\)

Correct answer:

\(\displaystyle 0\)

Explanation:

Remember the fundamental theorem of calculus!

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

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

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

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

Example Question #24 : Calculus Ii — Integrals

\(\displaystyle \int_{2}^{4}x^2{\mathrm{d} x}=?\)

Possible Answers:

\(\displaystyle 22\tfrac{1}{3}\)

\(\displaystyle 18\frac{2}{3}\)

\(\displaystyle 28\tfrac{2}{3}\)

\(\displaystyle 5\tfrac{1}{2}\)

\(\displaystyle 11\tfrac{3}{4}\)

Correct answer:

\(\displaystyle 18\frac{2}{3}\)

Explanation:

Remember the fundamental theorem of calculus!

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

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

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

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

Example Question #5 : Integrals

\(\displaystyle \int_{1}^{2}\frac{1}{x}{\mathrm{d} x}=?\)

Possible Answers:

\(\displaystyle 0.69\)

\(\displaystyle 0.75\)

\(\displaystyle 0.30\)

\(\displaystyle 0.81\)

\(\displaystyle 1.2\)

Correct answer:

\(\displaystyle 0.69\)

Explanation:

Remember the fundamental theorem of calculus!

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

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

\(\displaystyle \int \frac{1}{x}{\mathrm{d} x}=\ln{x}+c\)

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

Example Question #6 : Integrals

\(\displaystyle \int_{3}^{40}e^x{\mathrm{d} x}=?\)

Possible Answers:

\(\displaystyle 2.35\cdot 10^{19}\)

\(\displaystyle 2.35\cdot 10^{9}\)

\(\displaystyle 1.05\cdot 10^{17}\)

\(\displaystyle 2.35\cdot 10^{17}\)

\(\displaystyle 2.35\cdot 10^{15}\)

Correct answer:

\(\displaystyle 2.35\cdot 10^{17}\)

Explanation:

Remember the fundamental theorem of calculus!

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

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

\(\displaystyle \int e^x{\mathrm{d} x}=e^x+c\)

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

Example Question #7 : Integrals

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

Possible Answers:

\(\displaystyle -12\tfrac{2}{3}\)

\(\displaystyle 26\tfrac{2}{3}\)

\(\displaystyle 204\tfrac{2}{3}\)

\(\displaystyle 402\tfrac{2}{3}\)

\(\displaystyle 401\tfrac{1}{3}\)

Correct answer:

\(\displaystyle 402\tfrac{2}{3}\)

Explanation:

Remember the fundamental theorem of calculus!

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

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

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

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

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

Example Question #31 : Calculus Ii — Integrals

\(\displaystyle \int_3^5\frac{x+3}{x}{\mathrm{d} x}=?\)

Possible Answers:

\(\displaystyle 3.53\)

\(\displaystyle 5.67\)

\(\displaystyle 6.12\)

\(\displaystyle 3.21\)

\(\displaystyle 1.33\)

Correct answer:

\(\displaystyle 3.53\)

Explanation:

Remember the fundamental theorem of calculus!

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

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

 

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

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

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

Therefore:

\(\displaystyle \int\frac{x+3}{x}{\mathrm{d} x}=x+3\ln{x}+c\)

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

Example Question #9 : Integrals

\(\displaystyle \int_2^5\sqrt{x}\text{ }{\mathrm{d} x}=?\)

Possible Answers:

\(\displaystyle 7.52\)

\(\displaystyle 1.89\)

\(\displaystyle 9.34\)

\(\displaystyle 5.56\)

\(\displaystyle 2.59\)

Correct answer:

\(\displaystyle 5.56\)

Explanation:

Remember the fundamental theorem of calculus!

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

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

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

This means that:

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

 

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

Example Question #293 : Ap Calculus Ab

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

Possible Answers:

\(\displaystyle 3.9\)

\(\displaystyle 71.12\)

\(\displaystyle 8.7\)

\(\displaystyle 7.8\)

\(\displaystyle 10.8\)

Correct answer:

\(\displaystyle 7.8\)

Explanation:

Remember the fundamental theorem of calculus!

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

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

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

Remember to include the \(\displaystyle + c\) for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

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

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

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

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

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

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