Computing Definite Integrals — Question 7

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Question 7

Evaluate the definite integral ∫01(3x2+2)xdx.\int_{0}^{1} (3x^2+2)\sqrt{x}\,dx.

Original worksheet page 1: question and worked solution for 5-7-007
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Question 7 - Solution

Begin by rewriting the integrand using powers of xx: (3x2+2)x=3x5/2+2x1/2.(3x^2+2)\sqrt{x} = 3x^{5/2}+2x^{1/2}.

Now integrate term by term: ∫3x5/2dx=3⋅x7/27/2=67x7/2,\int 3x^{5/2}\,dx = 3\cdot\frac{x^{7/2}}{7/2} = \frac{6}{7}x^{7/2}, ∫2x1/2dx=2⋅x3/23/2=43x3/2.\int 2x^{1/2}\,dx = 2\cdot\frac{x^{3/2}}{3/2} = \frac{4}{3}x^{3/2}.

Combine the results: ∫(3x2+2)xdx=67x7/2+43x3/2.\int (3x^2+2)\sqrt{x}\,dx = \frac{6}{7}x^{7/2}+\frac{4}{3}x^{3/2}.

Evaluate from 00 to 11: [67x7/2+43x3/2]01=67+43.\left[\frac{6}{7}x^{7/2}+\frac{4}{3}x^{3/2}\right]_{0}^{1} = \frac{6}{7}+\frac{4}{3}.

Find a common denominator: 67+43=18+2821=4621.\frac{6}{7}+\frac{4}{3} = \frac{18+28}{21} = \frac{46}{21}.

4621\boxed{\frac{46}{21}}

Original worksheet page 2: question and worked solution for 5-7-007

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