Q. 74 - 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 

${\int }_a^b{x}^2 dx=\frac{1}{3}(b^3-a^3)$

Step-by-Step Solution

Answer

Proved that ${\int }_a^b{x}^2 dx=\frac{1}{3}(b^3-a^3)$

Step 1. Given information

The given integral ${\int }_a^b{x}^2 dx=\frac{1}{3}(b^3-a^3)$

Step 2. Prove that ∫ a b x 2   d x = 1 3 ( b 3 - a 3 ) using the fundamental theorem of calculas

$$\begin{gathered} {\int }_a^b{x}^2 dx={\left[\frac{x^3}{3}\right]}_a^b \\ =\frac{1}{3}\left(b^3\right)-\frac{1}{3}\left(a^3\right) [u\sin g fundamental theorem] \\ =\frac{1}{3}\left(b^3-a^3\right) \end{gathered}$$

Therefore,it is proved that ${\int }_a^b{x}^2 dx=\frac{1}{3}(b^3-a^3)$