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High School Math : Finding Definite Integrals

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Example Questions

Example Question #11 : Finding Integrals

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

Possible Answers:

$2\pi$

$\frac{\pi}{2}$

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

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

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

Now we can plug that equation into our FToC equation:

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

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

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

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

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

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Example Question #11 : Finding Integrals

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

Possible Answers:

$\frac{\pi}{2}$

$\pi$

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

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

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

Now we can plug that equation into our FToC equation:

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

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

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

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

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

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Example Question #12 : Finding Integrals

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

Possible Answers:

$2\pi$

$\pi$

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 $f(x)=\cos(2x)$ , we can't use the power rule. Instead we must use u-substituion. If $u = 2x, du = 2dx,and\ dx = \frac{du}{2}$ .

$\int f(x){\mathrm{d} x}=\int \cos(2x){\mathrm{d} x}= \int cos(u) \cdot \frac{1}{2}dx)= \frac{sin(u)}{2}+c=\frac{\sin(2x)}{2}+c$

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

Now we can plug that equation into our FToC equation:

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

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

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

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

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

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Example Question #13 : Finding Integrals

$\int_5^{10}\2x^2{\mathrm{d} x}=?$

Possible Answers:

$583\tfrac{1}{3}$

$58\tfrac{1}{3}$

750

$666\tfrac{2}{3}$

$58\tfrac{2}{3}$

Correct answer:

$583\tfrac{1}{3}$

Explanation:

Remember the fundamental theorem of calculus!

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

Since our f(x)=2x^2 , we can use the reverse power rule to find that the antiderivative is:

$\int2x^2{\mathrm{d} x}=\frac{2}{3}x^3+c$

Remember to include a for any integral or antiderivative taken!

Now we can plug that equation into our FToC equation:

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

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

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

$\int_5^{10} 2x^2{\mathrm{d} x}=(\frac{2}{3}(10)^3)-(\frac{2}{3}(5)^3)$

$\int_5^{10} 2x^2{\mathrm{d} x}=(666\tfrac{2}{3})-(88\tfrac{1}{3})$

$\int_5^{10} 2x^2{\mathrm{d} x}=583\tfrac{1}{3}$

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Example Question #14 : Finding Integrals

If n is a positive integer, find $\int_{0}^{2\pi}xsinxdx$ .

Possible Answers:

$-\frac{\pi }{n}$

$-2\pi$

$\frac{\pi }{n}$

$\frac{2\pi }{n}$

Correct answer:

$-2\pi$

Explanation:

We can find the integral using integration by parts, which is written as follows:

$\int udv=uv-\int vdu$

Let u=x and dv=sinxdx . We can get the box below:

$\begin{Bmatrix} u=x\\ du=dx\\ dv=sinxdx\\ v=-cosx+c \end{Bmatrix}$

Now we can write:

$\int_{0}^{2\pi}xsinxdx=\left |-xcosx+cx \right |_{0}^{2\pi }-\int_{0}^{2\pi} (-cosx+c)dx$

$=\left |-xcosx+cx+sinx-cx \right |_{0}^{2\pi }$

$=(-2\pi +0)-(0+0)$

$=-2\pi$

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Example Question #15 : Finding Integrals

$\int^5_25x+8{\mathrm{d} x}$ ?

Possible Answers:

102.5

76.5

128.5

64.25

Correct answer:

76.5

Explanation:

Remember the fundamental theorem of calculus! If g'(x)=f(x) , then $\int^b_af(x){\mathrm{d} x}=g(b)-g(a)$ .

Since we're given f(x) , we need to find the indefinite integral of the equation to get g(x) .

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

$\int 5x+8{\mathrm{d} x}=\frac{5x^{1+1}}{1+1}+\frac{8x^{0+1}}{0+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

$\int 5x+8{\mathrm{d} x}=\frac{5x^{2}}{2}+\frac{8x^{1}}{1}+c$

$\int 5x+8{\mathrm{d} x}=\frac{5}{2}x^2+8x +c$

Now we can plug that back in:

$\int^b_a 5x+8{\mathrm{d} x}=(\frac{5}{2}b^2+8b +c)-(\frac{5}{2}a^2+8a +c)$

Notice that the 's cancel out.

Plug in our given numbers.

$\int^5_2 5x+8{\mathrm{d} x}=(\frac{5}{2}(5)^2+8*5)-(\frac{5}{2}(2)^2+8*2)$

$\int^5_2 5x+8{\mathrm{d} x}=(\frac{5}{2}*25+40)-(\frac{5}{2}*4+16)$

$\int^5_2 5x+8{\mathrm{d} x}=(62.5+40)-(10+16)$

$\int^5_2 5x+8{\mathrm{d} x}=(102.5)-(26)$

$\int^5_2 5x+8{\mathrm{d} x}=76.5$

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Example Question #91 : Asymptotic And Unbounded Behavior

$\int_4^5 x^3+2x+5$ ?

Possible Answers:

106.25

53.5

100

206.25

306.25

Correct answer:

106.25

Explanation:

Remember the fundamental theorem of calculus! If g'(x)=f(x) , then $\int^b_af(x){\mathrm{d} x}=g(b)-g(a)$ .

Since we're given f(x) , we need to find the indefinite integral of the equation to get g(x) .

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent.

We're going to treat as 5x^0 , as anything to the zero power is one.

For this problem, that would look like:

$\int x^3+2x+5{\mathrm{d} x}=\frac{x^{3+1}}{3+1}+\frac{2x^{1+1}}{1+1}+\frac{5x^{0+1}}{0+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

$\int x^3+2x+5{\mathrm{d} x}=\frac{x^{4}}{4}+\frac{2x^{2}}{2}+\frac{5x^{1}}{1}+c$

$\int x^3+2x+5{\mathrm{d} x}=\frac{1}{4}x^4+x^2+5x+c$

Plug that back into the FTOC:

$\int_a^b x^3+2x+5{\mathrm{d} x}=(\frac{1}{4}b^4+b^2+5b+c)-(\frac{1}{4}a^4+a^2+5a+c)$

Notice that the 's cancel out.

Plug in our given values from the problem.

$\int_4^5 x^3+2x+5{\mathrm{d} x}=(\frac{1}{4}(5)^4+(5)^2+5(5))-(\frac{1}{4}(4)^4+(4)^2+5(4))$ $\int_4^5 x^3+2x+5{\mathrm{d} x}=(\frac{1}{4}*625+25+25)-(\frac{1}{4}*256+16+20)$

$\int_4^5 x^3+2x+5{\mathrm{d} x}=(156.25+25+25)-(64+16+20)$

$\int_4^5 x^3+2x+5{\mathrm{d} x}=(206.25)-(100)$

$\int_4^5 x^3+2x+5{\mathrm{d} x}=106.25$

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Example Question #11 : Integrals

$\int_4^8 6x^2+8{\mathrm{d} x}=?$

Possible Answers:

1088

160

928

1488

1248

Correct answer:

928

Explanation:

Use the Fundamental Theorem of Calculus: If g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ .

Therefore, we need to find the indefinite integral of our equation first.

To do that, we can use the anti-power rule or reverse power rule. We raise the exponent on the variables by one and divide by the new exponent.

For this problem, we'll treat as 8x^0 since anything to the zero power is one.

$\int 6x^2+8{\mathrm{d} x}=\frac{6x^{2+1}}{2+1}+\frac{8x^{0+1}}{0+1}+c$

Since the derivative of any constant is , when we take the indefinite integral, we add a to compensate for any constant that might be there.

From here we can simplify.

$\int 6x^2+8{\mathrm{d} x}=\frac{6x^{3}}{3}+\frac{8x^{1}}{1}+c$

$\int 6x^2+8{\mathrm{d} x}=2x^3+8x+c$

That means that $\int_a^b6x^2+8{\mathrm{d} x}=(2b^3+8b+c)-(2a^3+8a+c)$ .

Notice that the 's cancel out.

From here, plug in our numbers.

$\int_4^86x^2+8{\mathrm{d} x}=(2(8)^3+8(8))-(2(4)^3+8(4))$

$\int_4^86x^2+8{\mathrm{d} x}=(2*512+64)-(2*64+32)$

$\int_4^86x^2+8{\mathrm{d} x}=(1024+64)-(128+32)$

$\int_4^86x^2+8{\mathrm{d} x}=(1088)-(160)$

$\int_4^86x^2+8{\mathrm{d} x}=928$

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Example Question #12 : Integrals

$\int_2^{12} x+1{\mathrm{d} x}=?$

Possible Answers:

Correct answer:

Explanation:

Use the Fundamental Theorem of Calculus: If g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ .

Therefore, we need to find the indefinite integral of our equation first.

To do that, we can use the anti-power rule or reverse power rule. We raise the exponent on the variables by one and divide by the new exponent.

For this problem, we'll treat as 1x^0 since anything to the zero power is one.

$\int x+1{\mathrm{d} x}=\frac{x^{1+1}}{1+1}+\frac{1x^{0+1}}{0+1}+c$

Since the derivative of any constant is , when we take the indefinite integral, we add a to compensate for any constant that might be there.

From here we can simplify.

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

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

According to FTOC:

$\int_a^b x+1{\mathrm{d} x}=(\frac{1}{2}b^2+b+c)-(\frac{1}{2}a^2+a+c)$

Notice that the 's cancel out.

Plug in our given information and solve.

$\int_2^{12} x+1{\mathrm{d} x}=(\frac{1}{2}(12)^2+(12))-(\frac{1}{2}(2)^2+(2))$

$\int_2^{12} x+1{\mathrm{d} x}=(\frac{1}{2}*144+12)-(\frac{1}{2}*4+2)$

$\int_2^{12} x+1{\mathrm{d} x}=(72+12)-(2+2)$

$\int_2^{12} x+1{\mathrm{d} x}=(84)-(4)$

$\int_2^{12} x+1{\mathrm{d} x}=80$

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Example Question #11 : Finding Definite Integrals

$\int_3^512x^2+13x+4{\mathrm{d} x}=?$

Possible Answers:

861

178.5

Undefined

682.5

504

Correct answer:

504

Explanation:

Use the Fundamental Theorem of Calculus: If g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ .

Therefore, we need to find the indefinite integral of our equation first.

To do that, we can use the anti-power rule or reverse power rule. We raise the exponent on the variables by one and divide by the new exponent.

For this problem, we'll treat as 4x^0 since anything to the zero power is one.

$\int 12x^2+13x+4{\mathrm{d} x}=\frac{12x^{2+1}}{2+1}+\frac{13x^{1+1}}{1+1}+\frac{4x^{0+1}}{0+1}+c$

Since the derivative of any constant is , when we take the indefinite integral, we add a to compensate for any constant that might be there.

From here we can simplify.

$\int 12x^2+13x+4{\mathrm{d} x}=\frac{12x^{3}}{3}+\frac{13x^{2}}{2}+\frac{4x^{1}}{1}+c$

$\int 12x^2+13x+4{\mathrm{d} x}=4x^3+\frac{13}{2}x^2+4x+c$

According to FTOC:

$\int_a^b 12x^2+13x+4{\mathrm{d} x}=(4b^3+\frac{13}{2}b^2+4b+c)-(4a^3+\frac{13}{2}a^2+4a+c)$

Notice that the 's cancel out.

Plug in our given numbers and solve.

$\int_3^5 12x^2+13x+4{\mathrm{d} x}=(4(5)^3+\frac{13}{2}(5)^2+4(5))-(4(3)^3+\frac{13}{2}(3)^2+4(3))$

$\int_3^5 12x^2+13x+4{\mathrm{d} x}=((4*125)+(\frac{13}{2}*25)+20)-((4*27)+(\frac{13}{2}*(9))+12)$

$\int_3^5 12x^2+13x+4{\mathrm{d} x}=(500+162.5+20)-(108+58.5+12)$

$\int_3^5 12x^2+13x+4{\mathrm{d} x}=(682.5)-(178.5)$

$\int_3^5 12x^2+13x+4{\mathrm{d} x}=504$

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High School Math : Finding Definite Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #11 : Finding Integrals

Example Question #11 : Finding Integrals

Example Question #12 : Finding Integrals

Example Question #13 : Finding Integrals

Example Question #14 : Finding Integrals

Example Question #15 : Finding Integrals

Example Question #91 : Asymptotic And Unbounded Behavior

Example Question #11 : Integrals

Example Question #12 : Integrals

Example Question #11 : Finding Definite Integrals

All High School Math Resources

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