_{r=1}ⁿ₀^{π/2}(r+θ)²θ\,dθ =

Correct Answer is : n(n2+3n+3)3n (n² + 3n + 3)/33n(n2+3n+3)

Correct Answer is : n(n2+3n+3)3n (n² + 3n + 3)/33n(n2+3n+3)

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AP and TS EAMCET Engineering Model Paper 20

Question : 77 of 160

Marks: +1, -0

\sum\limits_{r=1}^{n}\int\limits_{0}^{\frac{\pi}{2}}(r+\sin\theta)^2\cos\theta\,d\theta =

Solution:

\sum\limits_{r=1}^{n}\int\limits_{0}^{\frac{\pi}{2}}(r+\sin\theta)^2\cos\theta\,d\theta

Let

\sin\theta = t \Rightarrow \cos\theta\,d\theta = dt

When

\theta = 0 \Rightarrow \sin 0 = 0 \Rightarrow t = 0

\theta = \frac{\pi}{2} \Rightarrow \sin\frac{\pi}{2} = 1 \Rightarrow t = 1

\sum\limits_{r=1}^{n}\int\limits_{0}^{1}(r+t)^2\,dt

= \int\limits_{0}^{1}(1+t)^2\,dt + \int\limits_{0}^{1}(2+t)^2\,dt + \dots

+ \int\limits_{0}^{1}(n+t)^2\,dt

= \left[\frac{(1+t)^3}{3}\right]_{0}^{1} + \left[\frac{(2+t)^3}{3}\right]_{0}^{1} + \dots

\dots + \left[\frac{(n+t)^3}{3}\right]_{0}^{1}

= \frac{2^3}{3} - \frac{1}{3} + \frac{3^3}{3} - \frac{2^3}{3} + \dots

+ \frac{n^3}{3} - \frac{(n-1)^3}{3} + \frac{(n+1)^3}{3} - \frac{n^3}{3}

= \frac{(n+1)^3}{3} - \frac{1}{3} = \frac{n^3 + 3n^2 + 3n + 1 - 1}{3}

= \frac{n (n^2 + 3n + 3)}{3}

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