Question

If the function $f:$ $[0,8]\to R$ is differentiable and $0<\alpha <1<\beta <2$

then ${\int }_0^{8}f\left(t\right)dt$ is equal to?

• A. $3\left[a^{3}f\right(a^2)+{\beta }^{2}f({\beta }^2\left)\right]$
• B. $3\left[a^{3}f\right(a)+{\beta }^{3}f(\beta \left)\right]$
• C. $3\left[a^{2}f\right(a^2)+{\beta }^{2}f({\beta }^3\left)\right]$
• D. $3\left[a^{2}f\right(a^3)+{\beta }^{2}f({\beta }^3\left)\right]$

Answer 1

Answer: (D)

$3\left[a^{2}f\right(a^3)+{\beta }^{2}f({\beta }^3\left)\right]$

Let $g\left(x\right)={\int }_0^{x^3}f\left(t\right)dt$

Now ${\int }_0^{8}f\left(t\right)dt=g\left(2\right)$

$=\left[g\right(2)-g(1\left)\right]+\left[g\right(1)-g(0\left)\right]$ ( $\because g\left(0\right)=0$)

$=\frac{g\left(2\right)-g\left(1\right)}{2-1}+\frac{g\left(1\right)-g\left(0\right)}{1-0}$

$=g^{\prime }\left(\beta \right)+g^{\prime }\left(\alpha \right)$ ....(1)

($\because f\left(x\right)$ is differentiable in $[0,8]\Rightarrow$ $g\left(x\right)$ is also differentiable in $[0,8]$. So we can use langrange's mean value theorem in the interval $[0,1]$ and $[1,2]$ )

Since, $g\left(x\right)={\int }_0^{x^3}f\left(t\right)dt$

$\Rightarrow g^{\prime }\left(x\right)=3x^{2}f\left(x^3\right)-0$ (By Leibnitz theorem)

$\Rightarrow g^{\prime }\left(\alpha \right)=3{\alpha }^{2}f\left({\alpha }^3\right)$

and $g^{\prime }\left(\beta \right)=3{\beta }^{2}f\left({\beta }^3\right)$

So, equation (1) becomes

${\int }_0^{8}f\left(t\right)dt=3\left[{\alpha }^{2}f\right({\alpha }^3)+{\beta }^{2}f({\beta }^3\left)\right]$