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High School Math : Calculus II — Integrals

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Example Question #41 : Calculus Ii — Integrals

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

Possible Answers:

128.5

64.25

102.5

76.5

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 #41 : Calculus Ii — Integrals

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

Possible Answers:

100

306.25

53.5

106.25

206.25

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 #43 : Calculus Ii — Integrals

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

Possible Answers:

1088

160

1488

928

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 #44 : Calculus Ii — 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 #45 : Calculus Ii — Integrals

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

Possible Answers:

Undefined

861

682.5

504

178.5

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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Example Question #46 : Calculus Ii — Integrals

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

Possible Answers:

399.8

413.6

427.4

Undefined

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 #47 : Calculus Ii — Integrals

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

Possible Answers:

71.25

241.5

49.5

142.5

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 #48 : Calculus Ii — Integrals

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

Possible Answers:

$250\tfrac{2}{3}$

$358\tfrac{1}{3}$

$259\tfrac{1}{3}$

$457\tfrac{1}{3}$

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 #49 : Calculus Ii — Integrals

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

Possible Answers:

$230\tfrac{2}{3}$

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

569

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

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 #50 : Calculus Ii — Integrals

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

Possible Answers:

$73\tfrac{5}{12}$

$179\tfrac{7}{12}$

$126\tfrac{2}{3}$

$53\tfrac{1}{4}$

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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High School Math : Calculus II — Integrals

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #41 : Calculus Ii — Integrals

Example Question #41 : Calculus Ii — Integrals

Example Question #43 : Calculus Ii — Integrals

Example Question #44 : Calculus Ii — Integrals

Example Question #45 : Calculus Ii — Integrals

Example Question #46 : Calculus Ii — Integrals

Example Question #47 : Calculus Ii — Integrals

Example Question #48 : Calculus Ii — Integrals

Example Question #49 : Calculus Ii — Integrals

Example Question #50 : Calculus Ii — Integrals

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