# The Area Bounded by the Curve y=x^3+1 and Line x=1

In this tutorial we shall find the area bounded by the curve $y = {x^3} + 1$, the x-axis and the line $x = 1$.

Since $y = 0$ at x-axis, for the points of intersection of the curve $y = {x^3} + 1$ with the x-axis, put $y = 0$. This implies that ${x^3} + 1 = 0$
$\begin{gathered} \Rightarrow \left( {x + 1} \right)\left( {{x^2} – x + 1} \right) = 0 \\ \Rightarrow x + 1 = 0,\,\,\,{x^2} – x + 1 = 0 \\ \Rightarrow x = – 1,\,\,\,x = \frac{{1 \pm \sqrt { – 3} }}{2} \\ \end{gathered}$

The curve cuts the x-axis only at $x = – 1$. The graph of the function $y = {x^3} + 1$ is shown in the given diagram. The required area of the shaded region.

The required area is given by the integral of the form
$A = \int\limits_{ – 1}^1 {ydx}$
$\begin{gathered} A = \int\limits_{ – 1}^1 {\left( {{x^3} + 1} \right)dx} \\ \Rightarrow A = \left| {\frac{{{x^4}}}{4} + x} \right|_{ – 1}^1 = \left( {\frac{{{1^4}}}{4} + 1} \right) – \left( {\frac{{{{\left( { – 1} \right)}^4}}}{4} + \left( { – 1} \right)} \right) \\ \Rightarrow A = \frac{1}{4} + 1 – \left( {\frac{1}{4} – 1} \right) = \frac{1}{4} + 1 – \frac{1}{4} + 1 \\ \end{gathered}$
$Area = 2$