Question

Use evaluation notation and the Fundamental Theorem of Calculus to prove Theorem 4.26:

${\int }_a^{b}f\left(x\right)dx={\left[\int f\left(x\right)dx\right]}_a^b$

Short answer

Ans:${\int }_a^b f\left(x\right)dx=\left[F\right(x){]}_a^b={\left[\int f\left(x\right)dx\right]}_a^b$

Step-by-step solution

Step 1. Given Information: 

${\int }_a^{b}f\left(x\right)dx={\left[\int f\left(x\right)dx\right]}_a^b$

Step 2. Prove:

$$\begin{gathered} ThefundamentaltheoremofCalculusis: \\ {\int }_a^{b}f\left(x\right)dx=\underset{n\to \infty }{\lim }\sum _{k=1}^n\left(F\left(x_k\right)-F\left(x_{k-1}\right)\right)=\underset{n\to \infty }{\lim }F\left(b\right)-F\left(a\right)=F\left(b\right)-F\left(a\right) \\ Where,Fisananti-derivativeoff \\ so, \\ {\int }_a^b f\left(x\right)dx=\left[F\right(x){]}_a^b={\left[\int f\left(x\right)dx\right]}_a^b \\ Henceproved. \end{gathered}$$