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High School Math : Finding Definite Integrals

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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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High School Math : Finding Definite Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #31 : Finding Definite Integrals

Example Question #51 : Calculus Ii — Integrals

Example Question #33 : Finding Definite Integrals

All High School Math Resources

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