Get the answer to "Evaluate the definite integral. 2∫01(y+1)1−ydy"

ka1leE

Answered question

2021-11-06

Evaluate the definite integral.

2 \int_{0}^{1} (y + 1) \sqrt{1 - y} d y

Mayme

Skilled2021-11-07Added 103 answers

Step 1: Given that
Evaluate the definite integral. Use a graphing utility to verify your result

2 \int_{0}^{1} (y + 1) \sqrt{1 - y} d y

Step 2: Solve

I = \int_{0}^{1} (y + 1) \sqrt{1 - y} d y

Substitute,
1−y=t
-dy=dt
Such that,

I = \int_{0}^{1} (1 - t + 1) \sqrt{t} \frac{d t}{- 1}

= \int_{0}^{1} 2 \sqrt{t} d t - \int_{0}^{1} t^{\frac{3}{2}} d t

= 2 {[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{1} - \frac{2}{5} {[t^{\frac{5}{2}}]}_{0}^{1}

= \frac{4}{3} {[{(1 - y)}^{\frac{3}{2}}]}_{0}^{1} - \frac{2}{5} {[{(1 - y)}^{\frac{3}{2}}]}_{0}^{1}

= - \frac{2}{5} + \frac{4}{3}

= \frac{20 - 6}{15}

= \frac{14}{15}

Therefore,

2 I = 2 \times \frac{14}{15} = \frac{28}{15}

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