Definite Integral

[jacks]

The value of \(\displaystyle \int_{0}^{1}4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx = \)



Asked in \(\bf{IIT\; JEE\; 2014}\)

[Papapetros Vaggelis]

\(\displaystyle{\begin{aligned}\dfrac{\mathrm{d}^2}{\mathrm{d}x^2}\,\left(1-x^2\right)^5&=\dfrac{\mathrm{d}}{\mathrm{d}x}\,\left(\dfrac{\mathrm{d}}{\mathrm{d}x}\,\left(1-x^2\right)^5\right)\\&=\dfrac{\mathrm{d}}{\mathrm{d}x}\,\left(-10\,x\,\left(1-x^2\right)^4\right)\\&=-10\,\left(1-x^2\right)^4+80\,x^2\,\left(1-x^2\right)^3\\&=-10\,\left(1-x^2\right)^4+80\,\left(1-x^2\right)^3-80\,\left(1-x^2\right)^4\\&=-90\,\left(1-x^2\right)^4+80\,\left(1-x^2\right)^3\end{aligned}}\)

Futhermore, \(\displaystyle{x\in\left[0,1\right]\implies 4\,x^3=\left(-2\,x^2\right)\,\left(-2\,x\right)=\left[2\,\left(1-x^2\right)-2\right]\,\left(1-x^2\right)'}\) .

So,

\(\displaystyle{\begin{aligned} \int_{0}^{1}4\,x^3\,\left\{\dfrac{\mathrm{d}^2}{\mathrm{d}x^2}\,\left(1-x^2\right)^5\right\}\,\mathrm{d}x&=\int_{0}^{1}\left(2\,\left(1-x^2\right)-2\right)\,\left(1-x^2\right)'\,\left[-90\,\left(1-x^2\right)^4+80\,\left(1-x^2\right)^3\right]\,\mathrm{d}x\\&=\int_{0}^{1}\left[2\,\left(1-x^2\right)-2\right]\,\left[-90\,\left(1-x^2\right)^4+80\,\left(1-x^2\right)^3\right]\,\mathrm{d}\,\left(1-x^2\right)\\&=\int_{0}^{1}\left(-180\,\left(1-x^2\right)^5+340\,\left(1-x^2\right)^4-160\,\left(1-x^2\right)^3\right)\,\mathrm{d}\,\left(1-x^2\right)\\&=\left[-30\,\left(1-x^2\right)^6+68\,\left(1-x^2\right)^5-40\,\left(1-x^2\right)^4\right]_{0}^{1}\\&=2\end{aligned}}\)

Question : What \(\displaystyle{\mathrm{IIT\,\, JEE\,\, 2014}}\) is ; Where \(\displaystyle{\mathrm{IIT\,\, JEE\,\, 2014}}\) takes place ;

[jacks]

Question : What \(\displaystyle{\mathrm{IIT\,\, JEE\,\, 2014}}\) is ; Where \(\displaystyle{\mathrm{IIT\,\, JEE\,\, 2014}}\) takes place ;

To Papapetros Vaggelis, Actually It is a paper for Engg. entrance exam for B.Tech(from IIT,NIT) in INDIA.
It Contain 3 papers, Physics,Chemistry and Maths.

[Tolaso J Kos]

Hello...
Vaggelis you don't have to evaluate the second derivative just the first one, because then we are going to apply \( \rm{IBP} \). See , let \( \displaystyle f(x)=\left (1-x^2 \right )^5 \), then \( \displaystyle f'(x)=-10x\left ( 1-x^2 \right )^4 \).
Thus the integral is rewritten as:
$$\begin{aligned}
\int_{0}^{1}4x^3f''(x)\,dx&= \left [ 4x^3f'(x) \right ]_0^1-\int_{0}^{1}12x^2f'(x) \,dx\\
&=-\int_{0}^{1}12x^2f'(x)\,dx=-\left [ 12x^2f(x) \right ]_0^1+\int_{0}^{1}24xf(x)\,dx \\
&= \int_{0}^{1}24x\left ( 1-x^2 \right )^5\,dx\\
&= \left [ -2\left ( 1-x^2 \right )^6 \right ]_0^1\\
&= 2
\end{aligned}$$

which evaluates the given integral.