Q. 63 - Calculus

Question

Prove that if $f$ has an antiderivative (say, $G$), then the function  $A\left(x\right)={\int }_0^{x}f\left(t\right)dt$ must also be an antiderivative of $f$. (Hint: Use the Fundamental Theorem of Calculus.)

Step-by-Step Solution

Answer

If $f$ has an antiderivative (say, $G$), then the function $A\left(x\right)={\int }_0^{x}f\left(t\right)dt$ must also be an antiderivative of $f$.

Step 1. Given information

We have to prove that If $f$ has an antiderivative (say, $G$), then the function $A\left(x\right)={\int }_0^{x}f\left(t\right)dt$ must also be an antiderivative of $f$.

Step 2. Proof of the given question.

Since $G$ is the antiderivative of $f$

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

So,

$$\begin{gathered} A\left(x\right)={\int }_0^{x}f\left(t\right)dt \\ ={\left[G\left(x\right)\right]}_0^x \\ =G\left(x\right)-G\left(0\right) \end{gathered}$$

$A$ and $G$ differ by a constant and have the same derivative, $f$.

Therefore, $A\left(x\right)={\int }_0^{x}f\left(t\right)dt$ must also be an antiderivative of $f$.