Q. 76 - 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 

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

Step-by-Step Solution

Answer

Proved that ${\int }_a^b\left(f\left(x\right)+g\left(x\right)\right) dx={\int }_a^{b}f\left(x\right)dx+{\int }_a^{b}g\left(x\right)dx$

Step 1. Given information

The given integral ${\int }_a^b\left(f\left(x\right)+g\left(x\right)\right) dx={\int }_a^{b}f\left(x\right)dx+{\int }_a^{b}g\left(x\right)dx$

Step 2. Prove that ∫ a b f x + g x   d x = ∫ a b f x d x + ∫ a b g x d x using the fundamental theorem of calculas

$$\begin{gathered} {\int }_a^b\left(f\left(x\right)+g\left(x\right)\right) dx={\left[f\left(x\right)+g\left(x\right)\right]}_a^b [u\sin g fundamental theorem] \\ =\left[\left(f\left(b\right)\right)-\left(f\left(a\right)\right)\right]+\left[\left(g\left(b\right)\right)-\left(g\left(a\right)\right)\right] \\ ={\left[F\left(x\right)\right]}_a^b+{\left[G\left(x\right)\right]}_a^b \left[F is an antiderivative of f G is an antiderivative of g\right] \\ ={\int }_a^{b}f\left(x\right)dx+{\int }_a^{b}g\left(x\right)dx \end{gathered}$$

Therefore,it is proved that ${\int }_a^b\left(f\left(x\right)+g\left(x\right)\right) dx={\int }_a^{b}f\left(x\right)dx+{\int }_a^{b}g\left(x\right)dx$