Topic : Integral of an algebraic expression
This is a simple problem illustrating definite integral.
Problem : Evaluate
Solution :
let x = 4 Sin Θ then dx/dΘ =4 Cos Θ or dx = 4 Cos Θ dΘ
As x = 4 Sin Θ, So x2 = (4 Sin Θ)2 = 16 Sin2 Θ
and when x = 0 ; 0 = 4 Sin Θ or Θ = Sin-10 or Θ = 0
when x = 4 ; 4 = 4 Sin Θ or Θ = Sin-14/4 or Θ = Sin-11 or Θ = π/2
Now we plug-in all the values,
![So, \int_{0}^{\pi/2}x^3\sqrt{16 - x^2} dx \\ = \int_{0}^{\pi/2}(4 Sin \theta)^3\sqrt{16 - 16 Sin^2 \theta} . 4 Cos \theta .d\theta \\ = \int_{0}^{\pi/2} 64 Sin^3 \theta)\sqrt{16(1 - Sin^2 \theta)} . 4 Cos \theta .d\theta \\= 256 \int_{0}^{\pi/2} Sin^3 \theta. 4 Cos \theta .Cos \theta .d\theta (as \sqrt{1 - Sin^2 \theta} = \sqrt{Cos^2 \theta} = Cos \theta) \\ = 1024 \int_{0}^{\pi/2} Sin^3 \theta .Cos^2 \theta . d\theta \\ =1024 \int_{0}^{\pi/2} Sin^2 \theta .Cos^2 \theta . Sin \theta.d\theta \\ = 1024 \int_{0}^{\pi/2} (1-Cos^2) .Cos^2 \theta . Sin \theta.d\theta \\ So, 1024 \int_{1}^{0}(1-t^2)t^2(-dt)\\ = 1024 \int_{1}^{0}-(t^2 -t^4)(-dt) \\ = 1024[\frac{t^3}{3}+\frac{t^5}{5}]_1^0 \\ = 1024 [-\frac{0^3}{3} + \frac{0^5}{5} - (-\frac{1^3}{3}-\frac{1^5}{5})] \\ = 1024(8/15) \\ = 546.13](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgmPj5AQ73rrljXyzr4E_6NkbHDvsewIewsmWA8Pn8DEdK_KkrarcCITMsBCaFfnrbffeOZqQ2q0Wk_YiuQMuBumI4jYfD-DqObpi6vabiV8n1PjY1MKdX3Qyn4jiKiceN6cjc9NN-DX3A/s1600-h/eval2.PNG)
Hope all the steps helped you to understand how to find derivative, when definite integral is given. For more help please do write to calculus help.
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