KCET · Maths · Area Under Curves
The area of the region bounded by \(y=2 x-x^{2}\) and the \(x\)-axis is
- A \(\frac{8}{3}\) sq unit
- B \(\frac{4}{3}\) sq unit
- C \(\frac{7}{3}\) sq unit
- D \(\frac{2}{3}\) sq unit
Answer & Solution
Correct Answer
(B) \(\frac{4}{3}\) sq unit
Step-by-step Solution
Detailed explanation
Given curve is
\[
y=2 x-x^{2}
\]
or
\[
(x-1)^{2}=-(y-1)
\]
The curve cut the \(x\)-axis at \((0,0)\) and \((2,0)\).
\(\therefore\) Required area \(=\int_{0}^{2} y \mathrm{dx}\)
\[
\begin{aligned}
&=\int_{0}^{2}\left(2 x-x^{2}\right) d x=\left[x^{2}-\frac{x^{3}}{3}\right]_{0}^{2} \\
&=4-\frac{8}{3}=\frac{4}{3} \text { sq unit }
\end{aligned}
\]
\[
y=2 x-x^{2}
\]
or
\[
(x-1)^{2}=-(y-1)
\]
The curve cut the \(x\)-axis at \((0,0)\) and \((2,0)\).
\(\therefore\) Required area \(=\int_{0}^{2} y \mathrm{dx}\)
\[
\begin{aligned}
&=\int_{0}^{2}\left(2 x-x^{2}\right) d x=\left[x^{2}-\frac{x^{3}}{3}\right]_{0}^{2} \\
&=4-\frac{8}{3}=\frac{4}{3} \text { sq unit }
\end{aligned}
\]
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