How do you integrate $\int x \sqrt{1 - x^{2}} d x$ $int xsqrt(1-x^2)dx$ from [0,1]?

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Answer 1 · 2016-12-04T14:59:29

$\frac{1}{3}$ $1/3$

If one recognises that the outside of the sqrt is a function of the sqrt differentiated we can proceed by inspection.

$\int_{0}^{1} x {(1 - x^{2})}^{\frac{1}{2}} d x$ $int_0^1x(1-x^2)^(1/2)dx$

guess $y = {(1 - x^{2})}^{\frac{3}{2}}$ $" " y=(1-x^2)^(3/2)$

$\frac{d}{d x} ({(1 - x^{2})}^{\frac{3}{2}}) = \frac{3}{2} \times (- 2 x) {(1 - 2 x)}^{\frac{1}{2}}$ $d/(dx)((1-x^2)^(3/2))=3/2xx(-2x)(1-2x)^(1/2)$

$= - 3 {(1 - 2 x)}^{\frac{1}{2}}$ $=-3(1-2x)^(1/2)$

$:.int_0^1x(1-x^2)^(1/2)dx=-1/3[(1-2x)^(3/2)]_0^1$

$-1/3{[(1-2x)^(3/2)]^1-[(1-2x)^(3/2)]_0}$

$-1/3((0)-1)$

$=1/3$

Answer 2 · 2016-12-04T15:08:47

$int_0^1 xsqrt(1-x^2)dx = 1/3$

Substitute:

$t = 1-x^2$
$dt = -2xdx$

$int_0^1 xsqrt(1-x^2)dx = -1/2int_1^0 sqrt(t) dt = 1/2int_0^1 sqrt(t) dt =$

$= 1/3 t^(3/2) |_(x=0)^(x=1) =1/3$

How do you integrate int xsqrt(1-x^2)dx∫x√1−x2dxint xsqrt(1-x^2)dx from [0,1]?