MHT CET · Maths · Area Under Curves
The area of the region bounded by the curve \(y=4 x-x^{2}\) and the \(x\) -axis is
- A \(\frac{16}{3}\) sq. units
- B \(\frac{32}{3}\) sq. units
- C 32 sq. units
- D 16 sq. units
Answer & Solution
Correct Answer
(B) \(\frac{32}{3}\) sq. units
Step-by-step Solution
Detailed explanation
We have \(y=4 x-x^{2}\)
When \(y=0\), we get \(x(4-x)=0 \Rightarrow x=0,4\)
Required area is shaded.
\(\begin{aligned}
\therefore A &=\int_{0}^{4}\left(4 x-x^{2}\right) d x \\
&=\left[\frac{4 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{4}=\left[2 x^{2}-\frac{x^{3}}{3}\right]_{0}^{4} \\
&=\left|2(16-0)-\frac{64-0}{3}\right|=\left|32-\frac{64}{3}\right| \\
&=\frac{32}{3}
\end{aligned}\)
When \(y=0\), we get \(x(4-x)=0 \Rightarrow x=0,4\)
Required area is shaded.
\(\begin{aligned}
\therefore A &=\int_{0}^{4}\left(4 x-x^{2}\right) d x \\
&=\left[\frac{4 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{4}=\left[2 x^{2}-\frac{x^{3}}{3}\right]_{0}^{4} \\
&=\left|2(16-0)-\frac{64-0}{3}\right|=\left|32-\frac{64}{3}\right| \\
&=\frac{32}{3}
\end{aligned}\)
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