MHT CET · Maths · Area Under Curves
The area of the region bounded by the curve \(y=2 x-x^2\) and \(\mathrm{X}\)-axis is
- A \(\frac{2}{3}\) sq. units
- B \(\frac{4}{3}\) sq. units
- C \(\frac{5}{3}\) sq. units
- D \(\frac{8}{3}\) sq. units
Answer & Solution
Correct Answer
(B) \(\frac{4}{3}\) sq. units
Step-by-step Solution
Detailed explanation
Point of intersection of curve \(y=2 x-x^2\) and \(x\) axis, is \(0=2 \mathrm{x}-\mathrm{x}^2 \Rightarrow \mathrm{x}(\mathrm{x}-2)=0 \Rightarrow \mathrm{x}=0,2\)
When \(\mathrm{x}=0, \mathrm{y}=0\) and when \(\mathrm{x}=2, \mathrm{y}=0\)
Refer figure
Required area is shaded
\(\begin{aligned}
& A=\int_0^2\left(2 x-x^2\right) d x \\
& =2\left[\frac{x^2}{2}\right]_0^2-\left[\frac{x^2}{3}\right]_0^2 \\
& =(4)-\left(\frac{8}{3}\right)=\frac{4}{3}
\end{aligned}\)
When \(\mathrm{x}=0, \mathrm{y}=0\) and when \(\mathrm{x}=2, \mathrm{y}=0\)
Refer figure
Required area is shaded
\(\begin{aligned}
& A=\int_0^2\left(2 x-x^2\right) d x \\
& =2\left[\frac{x^2}{2}\right]_0^2-\left[\frac{x^2}{3}\right]_0^2 \\
& =(4)-\left(\frac{8}{3}\right)=\frac{4}{3}
\end{aligned}\)
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