Q. 2 - Calculus

Question

Functions Defined by Integrals: Suppose $A\left(x\right)$ is the function that for each $x>0$ is equal to the area under the graph of $f\left(t\right)=t^2$ from 0 to x. 

Calculate $A\left(0\right),A\left(1\right),A\left(2\right),A\left(3\right) and A\left(4\right)$

Step-by-Step Solution

Answer

Values of 

$$\begin{gathered} A\left(0\right)=0 \\ A\left(1\right)=\frac{1}{3} \\ A\left(2\right)=\frac{8}{3} \\ A\left(3\right)=\frac{27}{3} \\ A\left(4\right)=\frac{64}{3} \end{gathered}$$

Step 1. Given information

The area under the graph of $f\left(t\right)=t^2$

Step 2. Finding the values of A ( 0 ) , A ( 1 ) , A ( 2 ) , A ( 3 )   a n d   A ( 4 )

Suppose $A\left(x\right)$is the function that for each $x>0$ is equal to the area under the graph of from 0 to x

given $f\left(t\right)=t^2$

The area under the graph of $j\left(t\right)=t$ can be witten in terms of a definate integral as below

$$\begin{gathered} A\left(x\right)={\int }_0^4 f\left(t\right) dt \\ =\frac{x^3}{3} \\ \sin ce, \\ A\left(x\right)=\frac{x^3}{3} \\ Then \\ A\left(0\right)=0 \\ A\left(1\right)=\frac{1}{3} \\ A\left(2\right)=\frac{8}{3} \\ A\left(3\right)=\frac{27}{3} \\ A\left(4\right)=\frac{64}{3} \end{gathered}$$