Find definite integral. ∫02x2(3x3+1)13dx

Question

Find definite integral.  ∫02x2(3x3+1)13dx

Jenny Sheppard · Accepted Answer

Step 1
The given integral can be solved by the method of substitution. The substitution that will be used is

3 x^{3} + 1 = u

. This gives

9 x^{2} d x = d u

or

x^{2} d x = \frac{d u}{9}

.
This substitution absorbs the

x^{2} d x

term into

\frac{d u}{9}

. Calculate the corresponding limits of the integration in terms of the new variable u.
Step 2
Limits of integration for x are from 0 to 2. New variable for integration is

u = 3 x^{3} + 1

. So the lower limit of integration in terms of new variable will be 1. Calculate the upper limit by substituting x=2.

u = 3 \cdot 2^{3} + 1

=3*8+1
=25
So, the integration with this substitution becomes

\int_{1}^{25} u^{\frac{1}{3}} \frac{d u}{9}

. Calculate this integral using the integral

\int x^{n} d x = \frac{x^{n + 1}}{n + 1}

.

\int_{1}^{25} u^{\frac{1}{3}} \frac{d u}{9} = \frac{1}{9} {(\frac{u^{\frac{1}{3} + 1}}{\frac{1}{3} + 1})}_{1}^{25}

= \frac{1}{9} \cdot \frac{3}{4} {(u^{\frac{4}{3}})}_{1}^{25}

= \frac{1}{12} (25^{\frac{4}{3}} - 1)

Hence, the given definite integral is equal to

\frac{1}{12} (25^{\frac{4}{3}} - 1)

Shawn Kim · Accepted Answer

∫02x2(3x3+1)1/3dx=∫x23x3+13dx =19∫u3du ∫u3du =3u434 19∫u3du =u4312 =(3x3+1)4312 ∫x23x3+13 =(3x3+1)4312+C

Vasquez · Accepted Answer

∫20x2(3x3+1)1/3dx∫x2×(3x3+1)1/3dx∫t139dt19×∫t13dt19×3tt3419×3(3x3+1)3x3+134(3x3+1)3x3+1312(3x3+1)3x3+1312|02(3×23+1)3×23+1312−(3×03+1)3×03+1312Answer:25253−112

Find definite integral. \int_{0}^{2}x^{2}(3x^{3}+1)^{1/3}dx

Answered question

Answer & Explanation

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