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High School Math : Finding Integrals

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Example Questions

Example Question #79 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

$\int_2^42x^4+\frac{1}{2}x{\mathrm{d} x}=?$

Possible Answers:

427.4

Undefined

399.8

413.6

Correct answer:

399.8

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.

To find the indefinite integral, we can use the reverse power rule: we raise the exponent by one and then divide by our new exponent.

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

Remember when taking the indefinite integral to include a to cover any potential constants.

Simplify.

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

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

Apply the FTOC:

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

Notice that the 's cancel out.

Plug in our given numbers and solve.

$\int_2^4 2x^4+\frac{1}{2}x{\mathrm{d} x}=(\frac{2}{5}(4)^5+\frac{1}{4}(4)^{2})-(\frac{2}{5}(2)^5+\frac{1}{4}(2)^{2})$

$\int_2^4 2x^4+\frac{1}{2}x{\mathrm{d} x}=(\frac{2}{5}*1024+\frac{1}{4}*16)-(\frac{2}{5}*32+\frac{1}{4}*4)$

$\int_2^4 2x^4+\frac{1}{2}x{\mathrm{d} x}=(409.6+4)-(12.8+1)$

$\int_2^4 2x^4+\frac{1}{2}x{\mathrm{d} x}=(413.6)-(13.8)$

$\int_2^4 2x^4+\frac{1}{2}x{\mathrm{d} x}=399.8$

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Example Question #2201 : High School Math

$\int_3^83x+12{\mathrm{d} x}=?$

Possible Answers:

71.25

49.5

142.5

241.5

192

Correct answer:

142.5

Explanation:

Use the Fundamental Theorem ofCcalculus. 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.

To find the indefinite integral, we can use the reverse power rule: we raise the exponent by one and then divide by our new exponent.

We are going to treat as 12x^0 since anything to the zero power is one.

$\int 3x+12{\mathrm{d} x}=\frac{3x^{1+1}}{1+1}+\frac{12x^{0+1}}{0+1}+c$

Remember when taking the indefinite integral to include a to cover any potential constants.

Simplify.

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

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

Plug that into our Fundamental Theorem of Calculus:

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

Notice that the 's cancel out.

Plug in our given numbers and solve.

$\int_3^8 3x+12{\mathrm{d} x}=(\frac{3}{2}(8)^2+12(8))-(\frac{3}{2}(3)^2+12(3))$

$\int_3^8 3x+12{\mathrm{d} x}=(\frac{3}{2}*64+12(8))-(\frac{3}{2}*9+12(3))$

$\int_3^8 3x+12{\mathrm{d} x}=(96+96)-(13.5+36)$

$\int_3^8 3x+12{\mathrm{d} x}=(192)-(49.5)$

$\int_3^8 3x+12{\mathrm{d} x}=142.5$

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Example Question #2202 : High School Math

$\int_3^55x^2+12x{\mathrm{d} x}=?$

Possible Answers:

$259\tfrac{1}{3}$

$358\tfrac{1}{3}$

$250\tfrac{2}{3}$

$457\tfrac{1}{3}$

Correct answer:

$259\tfrac{1}{3}$

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.

To find the indefinite integral, we use the reverse power rule. That means we raise the exponent on the variables by one and then divide by the new exponent.

$\int 5x^2+12x{\mathrm{d} x}=\frac{5x^{2+1}}{2+1}+\frac{12x^{1+1}}{1+1}+c$

Remember to include a when computing integrals. This is a place holder for any constant that might be in the new expression.

$\int 5x^2+12x{\mathrm{d} x}=\frac{5x^{3}}{3}+\frac{12x^{2}}{2}+c$

$\int 5x^2+12x{\mathrm{d} x}=\frac{5}{3}x^3+\frac{12}{2}x^2+c$

$\int 5x^2+12x{\mathrm{d} x}=\frac{5}{3}x^3+6x^2+c$

Now plug that back into the FTOC:

$\int_a^b 5x^2+12x{\mathrm{d} x}=(\frac{5}{3}b^3+6b^2+c)-(\frac{5}{3}a^3+6a^2+c)$

Notice that the 's cancel out.

Plug in our given numbers.

$\int_3^5 5x^2+12x{\mathrm{d} x}=(\frac{5}{3}(5)^3+6(5)^2)-(\frac{5}{3}(3)^3+6(3)^2)$

$\int_3^5 5x^2+12x{\mathrm{d} x}=(\frac{5}{3}(125)+6(25))-(\frac{5}{3}(27)+6(9))$

$\int_3^5 5x^2+12x{\mathrm{d} x}=(208\tfrac{1}{3}+150)-(45+54)$

$\int_3^5 5x^2+12x{\mathrm{d} x}=(358\tfrac{1}{3})-(99)$

$\int_3^5 5x^2+12x{\mathrm{d} x}=259\tfrac{1}{3}$

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Example Question #2203 : High School Math

$\int_0^4-8x^2+15=?$

Possible Answers:

$-230\tfrac{2}{3}$

$-110\tfrac{2}{3}$

$230\tfrac{2}{3}$

569

Correct answer:

$-110\tfrac{2}{3}$

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.

To find the indefinite integral, we use the reverse power rule. That means we raise the exponent on the variables by one and then divide by the new exponent.

$\int -8x^2+15{\mathrm{d} x}=\frac{-8x^{2+1}}{2+1}+\frac{15x^{0+1}}{0+1}+c$

Remember to include a when computing integrals. This is a place holder for any constant that might be in the new expression.

$\int -8x^2+15{\mathrm{d} x}=\frac{-8x^{3}}{3}+\frac{15x^{1}}{1}+c$

$\int -8x^2+15{\mathrm{d} x}=-\frac{8}{3}x^3+15x+c$

Plug that back into FTOC:

$\int_a^b -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}b^3+15b+c)-(-\frac{8}{3}a^3+15a+c)$

Notice that the 's cancel out.

Plug in our given numbers.

$\int_0^4 -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}(4)^3+15(4))-(-\frac{8}{3}(0)^3+15(0))$ $\int_0^4 -8x^2+15{\mathrm{d} x}=(-\frac{8}{3}(64)+(60))-(0+0)$

$\int_0^4 -8x^2+15{\mathrm{d} x}=(-170\tfrac{2}{3}+(60))$

$\int_0^4 -8x^2+15{\mathrm{d} x}=-110\tfrac{2}{3}$

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Example Question #2204 : High School Math

$\int_3^4x^3+2x^2+5=?$

Possible Answers:

$53\tfrac{1}{4}$

$179\tfrac{7}{12}$

$73\tfrac{5}{12}$

$126\tfrac{2}{3}$

Correct answer:

$73\tfrac{5}{12}$

Explanation:

The Fundamental Theorem of Calculus states that if g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ . Therefore, we need to find the indefinite integral of our given equation.

To find the indefinite integral, we can use the reverse power rule. Raise the exponent of the variable by one and then divide by that new exponent.

We're going to treat as 5x^0 .

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

Remember to include the when taking the integral to compensate for any constant.

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

Simplify.

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

Plug that into FTOC:

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

Notice that the 's cancel out.

Plug in our given numbers.

$\int_3^4 x^3+2x^2+5{\mathrm{d} x}=(\frac{1}{4}(4)^4+\frac{2}{3}(4)^3+5(4))-(\frac{1}{4}(3)^4+\frac{2}{3}(3)^3+5(3))$ $\int_3^4 x^3+2x^2+5{\mathrm{d} x}=(\frac{1}{4}(256)+\frac{2}{3}(64)+20)-(\frac{1}{4}(81)+\frac{2}{3}(27)+15)$

$\int_3^4 x^3+2x^2+5{\mathrm{d} x}=(64+42\tfrac{2}{3}+20)-(20\tfrac{1}{4}+18+15)$

$\int_3^4 x^3+2x^2+5{\mathrm{d} x}=(126\tfrac{2}{3})-(53\tfrac{1}{4})$

$\int_3^4 x^3+2x^2+5{\mathrm{d} x}=73\tfrac{5}{12}$

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Example Question #2205 : High School Math

$\int_2^44x^2+\frac{2}{3}x{\mathrm{d} x}=?$

Possible Answers:

$78\tfrac{1}{3}$

$102\tfrac{2}{3}$

308

$78\tfrac{2}{3}$

708

Correct answer:

$78\tfrac{2}{3}$

Explanation:

To find the definite integral, we can use the Fundamental Theorem of Calculus that states that 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 to start.

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

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

Remember to include a to cover any potential constant that might be in our new equation.

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

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

Plug that into FTOC:

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

Notice that the 's cancel out.

Plug in our given values.

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(4)^3+\frac{1}{3}(4)^2)-(\frac{4}{3}(2)^3+\frac{1}{3}(2)^2)$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(64)+\frac{1}{3}(16))-(\frac{4}{3}(8)+\frac{1}{3}(4))$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{256}{3}+\frac{16}{3})-(\frac{32}{3}+\frac{4}{3})$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{272}{3})-(\frac{36}{3})$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=\frac{236}{3}$

$\int_2^4 4x^2+\frac{2}{3}x{\mathrm{d} x}=78\tfrac{2}{3}$

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Example Question #2206 : High School Math

$\int_3^64x^2+\frac{2}{3}x{\mathrm{d} x}=?$

Possible Answers:

7.69

339

300

261

Correct answer:

261

Explanation:

To find the definite integral, we can use the Fundamental Theorem of Calculus which states that 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 to start.

To find the indefinite integral, we can use the reverse power rule. We raise the exponent of the variable by one and divide by our new exponent.

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

Remember to include a to cover any potential constant that might be in our new equation.

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

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

Plug that into FTOC:

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

Notice that the 's cancel out.

Plug in our given values.

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(6)^3+\frac{1}{3}(6)^2)-(\frac{4}{3}(3)^3+\frac{1}{3}(3)^2)$

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(\frac{4}{3}(216)+\frac{1}{3}(36))-(\frac{4}{3}(27)+\frac{1}{3}(9))$

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(288+12)-(36+3)$

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=(300)-(39)$

$\int_3^6 4x^2+\frac{2}{3}x{\mathrm{d} x}=261$

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Example Question #21 : Finding Definite Integrals

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

Possible Answers:

$2\pi$

Correct answer:

Explanation:

The fundamental theorem of calculus states that if f'(x)=g(x) , then $\int_a^bg(x){\mathrm{d} x}=f(b)-f(a)$ .

First, we need to find the indefinite integral of our given equation. Just like with the derivatives, the indefinite integrals or anti-derivatives of trig functions must be memorized.

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

Don't forget the to compensate for any potential constant!

Plug this in to our FTOC:

$\int_b^a \sin(x){\mathrm{d} x}=(-\cos(b)+c)-(-\cos(a)+c)$ .

Notice that the 's cancel out.

$\int_b^a \sin(x){\mathrm{d} x}=-\cos(b)+\cos(a)$ .

Now plug in the given values.

$\int_{-\pi}^{\pi} \sin(x){\mathrm{d} x}=-\cos(\pi)+\cos(-\pi)$

$\int_{-\pi}^{\pi} \sin(x){\mathrm{d} x}=0+0$

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

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Example Question #29 : Finding Integrals

$\int_2^8 4x^2+5x-3\: dx=?$

Possible Answers:

216

804

810

$818\tfrac{2}{3}$

270

Correct answer:

804

Explanation:

To solve for the definite integral, use the fundamental theorem of calculus. If g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ .

First we need to find the indefinite integral.

To find the indefinite integral of our given equation, we can use the reverse power rule: we raise the exponent by one and then divide by that new exponent.

$\int 4x^2+5x-3\: dx=\frac{4x^{2+1}}{2+1}+\frac{5x^{1+1}}{1+1}-\frac{3x^{0+1}}{0+1}+c$

Don't forget to include a to compensate for any constant!

$\int 4x^2+5x-3\: dx=\frac{4x^{3}}{3}+\frac{5x^{2}}{2}-\frac{3x^{1}}{1}+c$

$\int 4x^2+5x-3\: dx=\frac{4}{3}x^3+\frac{5}{2}x^2-3x+c$

Plug this into our first FTOC equation:

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

Notice that the 's cancel out.

$\int_a^b 4x^2+5x-3\: dx=(\frac{4}{3}b^3+\frac{5}{2}b^2-3b)-(\frac{4}{3}a^3+\frac{5}{2}a^2-3a)$

Plug in our given values.

$\int_2^8 4x^2+5x-3\: dx=(\frac{4}{3}(8)^3+\frac{5}{2}(8)^2-3(8))-(\frac{4}{3}(2)^3+\frac{5}{2}(2)^2-3(2))$

$\int_2^8 4x^2+5x-3\: dx=(682\tfrac{2}{3}+160-24)-(10\tfrac{2}{3}+10-6)$

$\int_2^8 4x^2+5x-3\: dx=(818\tfrac{2}{3})-(14\tfrac{2}{3})$

$\int_2^8 4x^2+5x-3\: dx=804$

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Example Question #30 : Finding Integrals

$\int_{3}^8 \frac{2}{3}x^3+7x^2-12x \ {\mathrm{d} x}=?$

Possible Answers:

1470.83

1711.82

1582.38

1512.83

1481.02

Correct answer:

1470.83

Explanation:

We can solve this problem using the Fundamental Theorem of Calculus:Iif g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ .

To use that equation for this problem, we need to find the indefinite integral of our given equation.

To find the indefinite integral of $\frac{2}{3}x^3+7x^2-12x$ , we can use the reverse power rule. To do this, we raise our exponent by one and then divide the variable by that new exponent.

$\int \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=\frac{\frac{2}{3}x^{3+1}}{3+1}+\frac{7x^{2+1}}{2+1}-\frac{12x^{1+1}}{1+1}+c$

Don't forget to include a to cover any constant!

$\int \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=\frac{\frac{2}{3}x^{4}}{4}+\frac{7x^{3}}{3}-\frac{12x^{2}}{2}+c$

$\int \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=\frac{2}{12}x^4+\frac{7}{3}x^3-6x^2+c$

$\int \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=\frac{1}{6}x^4+\frac{7}{3}x^3-6x^2+c$

Now we can plug that into the FTOC:

$\int_a^b \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=(\frac{1}{6}b^4+\frac{7}{3}b^3-6b^2+c)-(\frac{1}{6}a^4+\frac{7}{3}a^3-6a^2+c)$

Notice that the 's cancel out.

$\int_a^b \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=(\frac{1}{6}b^4+\frac{7}{3}b^3-6b^2)-(\frac{1}{6}a^4+\frac{7}{3}a^3-6a^2)$

Plug in our given values:

$\int_3^8 \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=(\frac{1}{6}(8)^4+\frac{7}{3}(8)^3-6(8)^2)-(\frac{1}{6}(3)^4+\frac{7}{3}(3)^3-6(3)^2)$

$\int_3^8 \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=(682\tfrac{2}{3}+1194\tfrac{2}{3}-384)-(13.5+63-54)$ $\int_3^8 \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=(1493\tfrac{1}{3})-(22.5)$

$\int_3^8 \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}\approx1470.83$

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High School Math : Finding Integrals

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #79 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #2201 : High School Math

Example Question #2202 : High School Math

Example Question #2203 : High School Math

Example Question #2204 : High School Math

Example Question #2205 : High School Math

Example Question #2206 : High School Math

Example Question #21 : Finding Definite Integrals

Example Question #29 : Finding Integrals

Example Question #30 : Finding Integrals

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