How do you find #int_-1^1x^2sqrt(4-x^2)dx# ?

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#int_-1^1x^2sqrt(4-x^2)dx#

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Answer 1 · 2018-06-22T12:06:27

# 2/3pi-sqrt3/2, or, 1/6(4pi-3sqrt3)#.

Let, #I=int_-1^1x^2sqrt(4-x^2)dx#.

Since the integrand is even, we have,

# I=2int_0^1x^2sqrt(4-x^2)dx#.

Let, #x=2sint,"so that, "dx=2costdt#.

#"Also, "x=0 rArr 2sint=0 rArr t=0, and, x=1 rArr t=pi/6#.

#:. I=2int_0^(pi/6)(4sin^2t)(sqrt(4-4sin^2t))(2cost)dt#,

#=8int_0^(pi/6)(4sin^2tcos^2t)dt#,

#=8int_0^(pi/6)sin^2 2tdt#,

#=8int_0^(pi/6){(1-cos(2xx2t))/2}dt#,

#=4int_0^(pi/6)(1-cos4t)dt#,

#=4[t-(sin4t)/4]_0^(pi/6)#,

#=[4t-sin4t]_0^(pi/6)#,

#=[{4*pi/6-sin(4*pi/6)}-0]#.

# rArr I=2/3pi-sqrt3/2=1/6(4pi-3sqrt3)#.

#color(green)("Enjoy Maths.!")#