# Find the area of the region bounded by the graphs of the given equations.

– $y \space = \space 4x \space + \space 5$   and $y \space = \space x^2$

The main objective of this question is to find the area of the bounded region for the given expression.

This question uses the concept of the area of the bounded region. The area of the bounded region can find by evaluating the definite integral.

We have to find the area of the bounded region.

So, given that:

$\space y \space = \space 4 x \space + \space 5$

$\space y \space = \space x^2$

Now for finding the intersecting point, we know that:

$\space 4 x \space + \space 5 \space = \space x^2$

$\space – 4 x \space – \space 5 \space + \space x^2 \space = \space 0$

$\space x^2 \space – \space 4 x \space – \space 5 \space = \space 0$

Solving the equation results in:

$\space x_1 \space = \space 5$

$\space x_2 \space = \space – \space 1$

By putting the values, we get:

$\space y \space = \space 4 x \space + \space 5$

$\space y \space = \space 4 ( 5 ) \space + \space 5$

$\space y \space = \space 2 0 \space + \space 5$

$\space y \space = \space 2 5$

Now putting $x_2$ value, results in:

$\space y \space = \space 4 ( – 1 ) \space + \space 5$

$\space y \space = \space – \space 4 \space + \space 5$

Thus:

$\space y \space = \space 1$

Thus, intersecting points are $(-1, \space 1)$ and $(5, \space 25)$  .

Now:

$\space A \space = \space \int_{-1}^{5} ( 4x \space + \space 5) \,dx \space – \space \int_{-1}^{5} ( x )^2 \,dx$

By simplifying, we get:

$\space = \space 78 \space – \space 42$

$\space = \space 36$

Thus:

$\space Area \space = \space 42$

The area for the given curve is:

$\space Area \space = \space 42$

## Example

Find the area of the bounded region by the two given curve equation.

$\space y \space = \space 5x \space + \space 6$

$\space y \space = \space x^2$

We have to find the area of the bounded region.

So, given that:

$\space y \space = \space 5 x \space + \space 6$

$\space y \space = \space x^2$

Now for finding the intersecting point, we know that:

$\space 5x \space + \space 6 \space = \space x^2$

$\space – 5 x \space – \space 6 \space + \space x^2 \space = \space 0$

$\space x^2 \space – \space 5 x \space – \space 6 \space = \space 0$

Solving the equation results in:

$\space x_1 \space = \space 6$

$\space x_2 \space = \space – \space 1$

By putting the values, we get:

$\space y \space = \space 5 x \space + \space 6$

$\space y \space = \space 4 ( 6 ) \space + \space 6$

$\space y \space = \space 2 4 \space + \space 6$

$\space y \space = \space 3 0$

Now putting $x_2$ value, results in:

$\space y \space = \space 5 ( – 1 ) \space + \space 6$

$\space y \space = \space – \space 5 \space + \space 6$

Thus:

$\space y \space = \space 1$

Now:

$\space A \space = \space \int_{-1}^{6} ( 5x \space + \space 6) \,dx \space – \space \int_{-1}^{6} ( x )^2 \,dx$

By simplifying, we get:

$\space = \space 57.2$

Thus:

$\space Area \space = \space 57.2$