Chapter 7.2, Problem 36E

To determine

To find: The area of the region.

Answer to Problem 36E

The area is: $A=\frac{15}{2}$

Explanation of Solution

Given information:

The region on or above the $x$ -axis bounded by the curve $y=4-x^2$

And line $y=3x$ .

Calculation:

The pink tint shades the portion of the region that needs to be evaluated. First, calculate the area of the region that is shown in the image as the product of the pink area and the purple triangle's area. By simply taking into account the following integral, obtain that area.

  

Finally, obtain the pink shaded region by deduct the area of the purple triangle from the area previously obtained. Since the purple triangle is a right-angled triangle, its area is represented as the cathetus product.

Purple triangle: $\frac{1\cdot 3}{2}=\frac{3}{2}$

Thus,

  $\begin{array}{c}\text{Pink area }=\text{ Area }-\text{ Purple triangle }\\ =9-\frac{3}{2}\\ =\frac{15}{2}\end{array}$

The graph is shown below:

  

Therefore, the required area of the region by subtracting the area of a triangular region from the area of a larger region is $\frac{15}{2}$ .