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High School Math : Calculus II — Integrals

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Example Question #51 : Calculus Ii — Integrals

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

Possible Answers:

$78\tfrac{2}{3}$

308

708

$102\tfrac{2}{3}$

$78\tfrac{1}{3}$

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 #52 : Calculus Ii — Integrals

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

Possible Answers:

300

261

7.69

339

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 #111 : Asymptotic And Unbounded Behavior

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

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 #21 : Finding Integrals

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

Possible Answers:

804

216

$818\tfrac{2}{3}$

270

810

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 #55 : Calculus Ii — Integrals

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

Possible Answers:

1512.83

1470.83

1711.82

1481.02

1582.38

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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Example Question #31 : Finding Definite Integrals

$\int_0^\pi 5x^4+\sin x\ {\mathrm{d} x}=?$

Possible Answers:

306.02

308.02

304.02

153

306.02

Correct answer:

308.02

Explanation:

To solve this problem we can 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.

Since we're adding terms, we take the integral or anti-derivative of each part separately. For 5x^4 , we can use the reverse power rule, which states that we raise the exponent of the variable by one and then divide by that new exponent. For sine, we use our trigonometric integral rules.

Remember, $\int \sin x \ {\mathrm{d} x}=-\cos x$ .

$\int 5x^4+\sin x\ {\mathrm{d} x}= (\frac{5x^{4+1}}{4+1})+(-\cos x)+c$

Don't forget to include a + c to account for any constants!

$\int 5x^4+\sin x\ {\mathrm{d} x}= (\frac{5x^{5}}{5})+(-\cos x)+c$

$\int 5x^4+\sin x\ {\mathrm{d} x}=x^5-\cos x +c$

Plug that into the FTOC:

$\int_a^b 5x^4+\sin x\ {\mathrm{d} x}=(b^5-\cos b +c)-(a^5-\cos a +c)$

Notice that the 's cancel out.

$\int_a^b 5x^4+\sin x\ {\mathrm{d} x}=(b^5-\cos b)-(a^5-\cos a)$

Plug in our given values:

$\int_0^\pi 5x^4+\sin x\ {\mathrm{d} x}=(\pi^5-\cos \pi)-(0^5-\cos (0))$

$\int_0^\pi 5x^4+\sin x\ {\mathrm{d} x}=(\pi^5-(-1))-(0-(1))$

$\int_0^\pi 5x^4+\sin x\ {\mathrm{d} x}=(\pi^5+1)-(-1))$

$\int_0^\pi 5x^4+\sin x\ {\mathrm{d} x}=\pi^5+2$

$\int_0^\pi 5x^4+\sin x\ {\mathrm{d} x}\approx 308.02$

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Example Question #51 : Calculus Ii — Integrals

$\int_{-3}^{3} x^4-3x^2+8\ {\mathrm{d} x}=?$

Possible Answers:

91.2

45.6

22.8

Correct answer:

91.2

Explanation:

To find the definite integral, we can use the Fundamental Theorem of Calculus. This states that if g'(x)=f(x) , then $\int_a^bf(x)=g(b)-g(a)$ .

To use the FToC, we need to find our indefinite integral of our given equation.

To find the indefinite integral, or anti-derivative, we can use the reverse power rule. We raise the exponent of each variable by one and divide by that new exponent.

$\int x^4-3x^2+8\ {\mathrm{d} x}=(\frac{x^{4+1}}{4+1})-(\frac{3x^{2+1}}{2+1})+(\frac{8x^{0+1}}{0+1})+c$

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

Simplify.

$\int x^4-3x^2+8\ {\mathrm{d} x}=(\frac{x^{5}}{5})-(\frac{3x^{3}}{3})+(\frac{8x^{1}}{1})+c$

$\int x^4-3x^2+8\ {\mathrm{d} x}=\frac{1}{5}x^5-x^3+8x+c$

We can now plug that into FToC!

$\int_a^b x^4-3x^2+8\ {\mathrm{d} x}=(\frac{1}{5}b^5-b^3+8b+c)-(\frac{1}{5}a^5-a^3+8a+c)$

Notice that the 's cancel out.

$\int_a^b x^4-3x^2+8\ {\mathrm{d} x}=(\frac{1}{5}b^5-b^3+8b)-(\frac{1}{5}a^5-a^3+8a)$

Plug in our given values.

$\int_{-3}^{3} x^4-3x^2+8\ {\mathrm{d} x}=(\frac{1}{5}(3)^5-(3)^3+8(3))-(\frac{1}{5}(-3)^5-(-3)^3+8(-3))$ $\int_{-3}^{3} x^4-3x^2+8\ {\mathrm{d} x}=(\frac{1}{5}(243)-(27)+(24))-(\frac{1}{5}(-243)-(-27)+(-24))$ $\int_{-3}^{3} x^4-3x^2+8\ {\mathrm{d} x}=((48.6)-(27)+(24))-((-48.6)-(-27)+(-24))$

$\int_{-3}^{3} x^4-3x^2+8\ {\mathrm{d} x}=(45.6)-(-45.6)$

$\int_{-3}^{3} x^4-3x^2+8\ {\mathrm{d} x}=91.2$

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Example Question #33 : Finding Definite Integrals

$\int^6_3 x^2+5x\ dx =$

Possible Answers:

Undefined

130.5

31.5

193.5

Correct answer:

130.5

Explanation:

Remember the Rundamental Theorem of Calculus: If g'(x)=f(x) , then $\int^b_af(x){\mathrm{d} x}=g(b)-g(a)$ .

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:

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

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

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

Now we can plug that back into the problem.

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

Notice that the 's cancel out. Plug in the values given in the problem:

$\int^6_3 x^2+5x{\mathrm{d} x}=(\frac{1}{3}(6)^3+\frac{5}{2}(6)^2)-(\frac{1}{3}(3)^3+\frac{5}{2}(3)^2)$ $\int^6_3 x^2+5x{\mathrm{d} x}=(\frac{1}{3}*216+\frac{5}{2}*36)-(\frac{1}{3}*27+\frac{5}{2}*9)$

$\int^6_3 x^2+5x{\mathrm{d} x}=(72+90)-(9+22.5)$

$\int^6_3 x^2+5x{\mathrm{d} x}=(162)-(31.5)$

$\int^6_3 x^2+5x{\mathrm{d} x}=130.5$

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Example Question #61 : Asymptotic And Unbounded Behavior

$\int \frac{3}{x}=$

Possible Answers:

$3\ln \left | x \right |+C$

$3x^{-2}+C$

$-3x^{-2}+C$

$3x^{-1}+C$

$-3e^{-x}$

Correct answer:

$3\ln \left | x \right |+C$

Explanation:

The integral of $\frac{1}{x}$ is $\ln \left | x \right |+C$ . The constant 3 is simply multiplied by the integral.

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Example Question #2 : Finding Indefinite Integrals

$\int cos(x)\sin(x){\mathrm{d} x}=?$

Possible Answers:

$\cos^2(x)sin^2(x)+c$

$\sin(x)\cos(x)+c$

$-(\frac{1}{2}\cos^2(x)+c)$

$\frac{\cos(x)}{\sin(x)}+c$

$\frac{\sin(x)}{\cos(x)}+c$

Correct answer:

$-(\frac{1}{2}\cos^2(x)+c)$

Explanation:

To integrate $\cos(x)\sin(x)$ , we need to get the two equations in terms of each other. We are going to use "u-substitution" to create a new variable, , which will equal $\cos(x)$ .

Now, if $u=\cos(x)$ , then

$\frac{\mathrm{d}u }{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-sin(x)$

Multiply both sides by $\mathrm d {x}$ to get the more familiar:

${\mathrm{d} u}=-\sin(x)\mathrm{d}x$

Note that our $\mathrm d{u}=-\sin(x)$ , and our original equation was asking for a positive $\sin(x)$ .

That means if we want $\int\cos(x)\sin(x)$ in terms of , it looks like this:

$\int u\cdot (-\mathrm d{u}})$

Bring the negative sign to the outside:

$-\int u\mathrm d{u}$ .

We can use the power rule to find the integral of :

$-(\frac{1}{2}u^2+c)$

Since we said that $u=\cos(x)$ , we can plug that back into the equation to get our answer:

$-(\frac{1}{2}\cos^2(x)+c)$

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High School Math : Calculus II — Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #51 : Calculus Ii — Integrals

Example Question #52 : Calculus Ii — Integrals

Example Question #111 : Asymptotic And Unbounded Behavior

Example Question #21 : Finding Integrals

Example Question #55 : Calculus Ii — Integrals

Example Question #31 : Finding Definite Integrals

Example Question #51 : Calculus Ii — Integrals

Example Question #33 : Finding Definite Integrals

Example Question #61 : Asymptotic And Unbounded Behavior

Example Question #2 : Finding Indefinite Integrals

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