Q. 20 - Calculus

Question

Are area accumulation functions the “inverse” of differentiation? In other words, does one undo the other? Simplify each of the following to answer this question:

$$\begin{gathered} \left(a\right) {\int }_a^x{f}^{\text{'}}\left(t\right)dt \\ \left(b\right) \frac{d}{dx}{\int }_a^{x}f\left(t\right)dt \end{gathered}$$

Step-by-Step Solution

Answer

Part $\left(a\right)$: The simplified expression for ${\int }_a^x{f}^{\text{'}}\left(t\right)dt$ is, $f\left(x\right)-f\left(a\right)$.

Part $\left(b\right)$: The simplified expression for $\frac{d}{dx}{\int }_a^{x}f\left(t\right)dt$ is, $f\left(x\right)$.

Part a Step 1 . Given information

${\int }_a^x{f}^{\text{'}}\left(t\right)dt$.

Part a Step 2 . The objective is, we need to simplify the given expression.

Now, $\frac{d}{dx}f\left(x\right)=f^{*}\left(x\right)$.

So,

$$\begin{gathered} {\int }_a^x{f}^{\text{'}}\left(t\right)dt=\left[f\right(t){]}_a^x \\ =f\left(x\right)-f\left(a\right) \end{gathered}$$

Part b Step 1 . Given information

$\frac{d}{dx}{\int }_a^{x}f\left(t\right)dt$.

Part b Step 2 . The objective is, we need to simplify the given expression.

Now, $F\left(x\right)=\int f\left(x\right)$             [ $F\left(x\right)$ is an antiderivative of $f\left(x\right)$]

So,

$$\begin{gathered} \frac{d}{dx}{\int }_a^{v}f\left(t\right)dt=\frac{d}{dx}\left[F\right(t){]}_a^x \\ =\frac{d}{dx}[F\left(x\right)-F\left(a\right)] \end{gathered}$$

                      $=\frac{d}{dx}F\left(x\right)$  [ $F\left(a\right)$ is a constant and derivative of the constant is $0$]

                       $=f\left(x\right)$