Q. 73 - Calculus

Question

Use the Fundamental Theorem of Calculus to give alternative proofs of the integration facts shown in Exercises 72–76. You may assume that all functions here are integrable 

${\int }_a^{b}x dx=\frac{1}{2}(b^2-a^2)$

Step-by-Step Solution

Answer

Proved that ${\int }_a^{b}x dx=\frac{1}{2}(b^2-a^2)$

Step 1. Given information

The given integral ${\int }_a^{b}x dx=\frac{1}{2}(b^2-a^2)$

Step 2. Prove that ∫ a b x   d x = 1 2 ( b 2 - a 2 ) using the fundamental theorem of calculas

$$\begin{gathered} {\int }_a^{b}x dx={\left[\frac{x^2}{2}\right]}_a^b \\ =\frac{1}{2}\left(b^2\right)-\frac{1}{2}\left(a^2\right) \\ =\frac{1}{2}\left(b^2-a^2\right) \\ =\frac{1}{2}(b^2-a^2) \end{gathered}$$

Therefore,it is proved that ${\int }_a^{b}x dx=\frac{1}{2}(b^2-a^2)$