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

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

Short 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-by-step solution

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 ∫abfx+gx dx=∫abfxdx+∫abgxdxusing the fundamental theorem of calculas

$$\begin{gathered} {\int }_a^b\left(f\left(x\right)+g\left(x\right)\right)dx={\left[f\left(x\right)+g\left(x\right)\right]}_a^b[u\sin gfundamentaltheorem] \\ =\left[\left(f\left(b\right)\right)-\left(f\left(a\right)\right)\right]+\left[\left(g\left(b\right)\right)-\left(g\left(a\right)\right)\right] \\ ={\left[F\left(x\right)\right]}_a^b+{\left[G\left(x\right)\right]}_a^b\left[FisanantiderivativeoffGisanantiderivativeofg\right] \\ ={\int }_a^{b}f\left(x\right)dx+{\int }_a^{b}g\left(x\right)dx \end{gathered}$$

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