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, so 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 x-axis only at x =  - 1. The graph of the given function y = {x^3} + 1 as shown in the given diagram. The required area of the shaded region.


area-bounded-curve-x-1

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


Which shows that the area under the curve.

Comments

comments