Key to "Use geometry and properties of integrals to evaluate..."

Tabansi

Answered question

2021-10-14

Use geometry and properties of integrals to evaluate the following definite integrals.

\int_{0}^{4} \sqrt{8 x - x^{2}} d x

Asma Vang

Skilled2021-10-15Added 93 answers

Complete the square:

(8 x - x^{2}) = - (x^{2} - 8 x + 16 - 16)

\Rightarrow (8 x - x^{2}) = - {(x - 4)}^{2} + 16

Substitute this integral, we get:

\int_{0}^{4} \sqrt{8 x - x^{2}} d x = \int_{0}^{4} \sqrt{- {(x - 4)}^{2} + 16} d x

Apply u-substitution
u=x-4

\Rightarrow d u = d x

for x=0,u=-4;x=4,u=0
Thus the integral in terms of u is given by:

\int_{0}^{4} \sqrt{8 x - x^{2}} d x = \int_{- 4}^{0} \sqrt{- u^{2} + 16} d u

Use trigonometric substitution

u = 4 \sin (v)

\Rightarrow d u = 4 \cos (v)

for

u = - 4, v = \frac{- π}{2}; u = 0, v = 0

The integral in terms of v is:

\int_{- \frac{π}{2}}^{0} \sqrt{- 16 \sin^{2} v + 16} (4 \cos v) d v

\Rightarrow 4 \int_{- \frac{π}{2}}^{0} \sqrt{16 (1 - \sin^{2} v)} \cos v d v

\Rightarrow 4 \int_{- \frac{π}{2}}^{0} \sqrt{16 \cos^{2} v} \cos v d v

\Rightarrow 4 \int_{- \frac{π}{2}}^{0} 4 \cos v \cos v d v

\Rightarrow 16 \int_{- \frac{π}{2}}^{0} \cos^{2} v d v

\Rightarrow 16 \int_{- \frac{π}{2}}^{0} \frac{1 + \cos (2 v)}{2} d v

\Rightarrow 8 {[v + \frac{1}{2} \sin (2 v)]}_{- \frac{π}{2}}^{0}

\Rightarrow 8 [(0 + 0) - (- \frac{π}{2} + \frac{1}{2} \sin (- π))]

\Rightarrow 4 π

Answer:

4 π

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