Question

If the function $f:[0,16]\to R$ is differentiable. If $0<\alpha <1$ and $1<\beta <2$, then ${\int }_0^{18}f\left(t\right)dt$ is equal to

• A. $4\left({\alpha }^{3}f\right({\alpha }^4)-{\beta }^{3}f({\beta }^4\left)\right)$
• B. $4\left({\alpha }^{3}f\right({\alpha }^4)+{\beta }^{3}f({\beta }^4\left)\right)$
• C. $4\left({\alpha }^{4}f\right({\alpha }^3)+{\beta }^{4}f({\beta }^3\left)\right)$
• D. $4\left({\alpha }^{2}f\right({\alpha }^2)-{\beta }^{2}f({\beta }^2\left)\right)$

Answer 1

The correct option is B $4\left({\alpha }^{3}f\right({\alpha }^4)+{\beta }^{3}f({\beta }^4\left)\right)$

Given : $f:[0,16]\to R0<\alpha <1\&1<\beta <2$

$g\left(x\right)={\int }_0^{16}f\left(t\right)dt=g\left(2\right)=\frac{g\left(2\right)-g\left(1\right)}{2-1}+\frac{g\left(1\right)-g\left(0\right)}{1-0}{\int }_0^{16}f\left(t\right)dt=g^{\prime }\left(\alpha \right)+g^{\prime }\left(\beta \right)=4{\alpha }^{3}f\left({\alpha }^4\right)+4{\beta }^{3}f\left({\beta }^4\right){\int }_0^{16}f\left(t\right)dt=4\left[{\alpha }^{3}f\right({\alpha }^4)+{\beta }^{3}f({\beta }^4\left)\right]$

Hence the correct answer is $4\left[{\alpha }^{3}f\right({\alpha }^4)+{\beta }^{3}f({\beta }^4\left)\right]$