Question

Let $f:R\to R$ be a differentiable function and $f\left(1\right)=4$. Then the value of ${\int }_4^{f\left(x\right)}\frac{2t}{x-1}dt$

Answer 1

Given $b:R\to R$ a differentiable function & $b\left(1\right)=4$

$\underset{x\to 1}{lim}{\int }_4^{b\left(x\right)}\frac{2t}{x-1}dt=\underset{x\to 1}{lim}{\left[\frac{t^2}{x-1}\right]}^{b\left(x\right)}$

$=\underset{x\to 1}{lim}\left(\frac{\left(b\right(x){)}^2-16}{x-1}\right)=\underset{x\to 1}{lim}\frac{b\left(x\right)-4}{x-1}.\underset{x\to 1}{lim}b\left(x\right)+4$

Using $b\left(1\right)=4$

$=b\left(1\right)\left(b\right(1)+4)=8b^{\prime }\left(1\right)$