#### Find the area of the region bounded by the curve $y = x^3$ and y = x + 6 and x = 0.

Let’s start with a rough plot of the curve $y = x\textsuperscript{3}$ along with the lines y = x + 6 and x = 0

When x = 0, it means Y axis

Questions says to find the area between curve and the line and Y axis

First solve the y = x + 6 and  $y = x\textsuperscript{3}$, in order to find the interaction point
$\\\text { Put } y=x^{3} \text { in } y=x+6 \\ \Rightarrow x^{3}=x+6 \\ \Rightarrow x^{3}-x-6=0 \\ \\$

For checking is 0, 1, 2 satisfy this cubic, shows 2 is one factor, therefore x – 2 is a factor.

Solving the equation,

$\\ \Rightarrow(x-2)\left(x^{2}+2 x+3\right)=0 \\ \text{Observe that } x^{2}+2 x+3 \text{ don't have real roots} \\$

Therefore, x = 2

Substituting this x = 2 in y = x + 6, y = 8

Therefore, curves intersect at (2, 8)

The area bounded will be

item Area bounded = area by $y = x\textsuperscript{3}$ on Y axis – area by y = x + 6 on Y axis   $\ldots (1) \\$

For finding area under $y = x\textsuperscript{3} \\$

$\Rightarrow x=\sqrt[3]{y} \\$

Integrate the equation from 0 to 8
$\\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\int_{0}^{8} \mathrm{y} \frac{1}{3} \mathrm{dy} \\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\left[\frac{\mathrm{y}^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]_{0}^{8} \\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\left[\frac{\mathrm{y}^{\frac{4}{3}}}{\frac{4}{3}}\right]_{0}^{8} \\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\frac{3}{4}\left[\mathrm{y}^{\frac{4}{3}}\right]_{0}^{8} \\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\frac{3}{4}[8 \overline{4/3}-0] \\ \\ \\$$\\\Rightarrow \int_{0}^{8} x d y=\frac{3}{4}\left(2^{3}\right)^{\frac{4}{3}} \\ \Rightarrow \int_{0}^{8} x d y=\frac{3}{4} 2^{4} \\ \Rightarrow \int_{0}^{8} x d y=\frac{3}{4} 16 \\ \Rightarrow \int_{0}^{8} x d y=12 \\ \\$

Now, finding area under y = x + 6

For finding the area from 6 to 8 because line passes through Y axis at 6 and extends upto 8, the point where curve and line intersects.

$\item X = y - 6 \\$

Integrating from 6 to 8
$\\ \Rightarrow \int_{6}^{8} x d y=\int_{6}^{8}(y-6) d y \\ \Rightarrow \int_{6}^{8} x d y=\left[\frac{y^{2}}{2}-6 y\right]_{6}^{8} \\ \Rightarrow \int_{6}^{8} x d y=\left[\left(\frac{8^{2}}{2}-6(8)\right)-\left(\frac{6^{2}}{2}-6(6)\right)\right] \\ \Rightarrow \int_{6}^{8} x d y=[32-48-18+36] \\ \Rightarrow \int_{6}^{8} x d y=2 \\ \\$

Using the equation (1)

Area bound was found to be 12 – 2 = 10 $unit\textsuperscript{2 }\\$

Therefore, the area was found to be 10 $unit\textsuperscript{2 }\\$