MHT CET · Maths · Area Under Curves
The area bounded by the parabola \(y=x^2\) and the line \(y=x\) is
- A \(\frac{1}{2}\) sq. units
- B \(\frac{1}{3}\) sq. units
- C \(\frac{2}{3}\) sq. units
- D \(\frac{1}{6}\) sq. units
Answer & Solution
Correct Answer
(D) \(\frac{1}{6}\) sq. units
Step-by-step Solution
Detailed explanation
The required area is shaded
The point of intersection of the curves are \(x^2=x \Rightarrow x(x-1)=0\) i.e. \(O(0,0)\) and \(P(1,1)\)
\(\begin{aligned} & \therefore A=\int_0^1\left(x-x^2\right) d x \\ & =\left[\frac{x^2}{2}\right]_0^1-\left[\frac{x^3}{3}\right]_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\end{aligned}\)
The point of intersection of the curves are \(x^2=x \Rightarrow x(x-1)=0\) i.e. \(O(0,0)\) and \(P(1,1)\)
\(\begin{aligned} & \therefore A=\int_0^1\left(x-x^2\right) d x \\ & =\left[\frac{x^2}{2}\right]_0^1-\left[\frac{x^3}{3}\right]_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\end{aligned}\)
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