Chapter 11.3, Problem 6E

To determine

Find the area of the curve $y=x^3$ from $x=0\text{to}x=4$ and above the x − axis.

Answer to Problem 6E

The approximate area under the curve is 62 sq units.

Explanation of Solution

Given:

Find the area of the curve $y=x^3$ from $x=0\text{to}x=4$ and above the x − axis.

Concept Used:

It is like sum of areas of 4 rectangles

The graph: $y=x^3$

  

Calculation:

It is like sum of areas of 4 rectangles.

We find the area under the curve $y=x^3$ from $x=0\text{to}x=4$ ,

According to the graph there are four rectangles.

First rectangle from $x=0\text{to}x=1$

Second rectangle from $x=1\text{to}x=2$

Third rectangle from $x=2\text{to}x=3$

Fourth rectangle from $x=3\text{to}x=4$

Width of each rectangle is 1 unit.

We use the midpoint of each rectangle.

Now find the height of each rectangle:

So height of first rectangle would be $x=0.5\Rightarrow {(0.5)}^3=0.125$

Height of second rectangle would be $x=1.5\Rightarrow {(1.5)}^3=3.375$

Height of third rectangle would be $x=2.5\Rightarrow {(2.5)}^3=15.625$

Height of fourth rectangle would be $x=3.5\Rightarrow {(3.5)}^3=42.875$

The area of 1^st, 2^nd, 3^rd and 4^th rectangles are : $0.125,3.375,15.625,42.875$ sq unit.

So the approximate area under the curve = sum of all 4 rectangles

So the approximate area under the curve: $0.125+3.375+15.625+42.875=62$ sq unit.

Thus, the approximate area under the curve is 62 sq units.