Question:
The area of the region bounded by the \(x\)-axis, the line \(x=4\) and the curve
\[
f(x)=
\begin{cases}
x^2, & 0\le x\le 1\\
\sqrt{x}, & x\gt 1
\end{cases}
\]
is
The area of the region bounded by the \(x\)-axis, the line \(x=4\) and the curve
\[
f(x)=
\begin{cases}
x^2, & 0\le x\le 1\\
\sqrt{x}, & x\gt 1
\end{cases}
\]
is
Show Hint
For piecewise functions:
\[
\int_a^b f(x)\,dx
=
\int_a^c f_1(x)\,dx
+
\int_c^b f_2(x)\,dx
\]
where \(c\) is the point at which the definition changes.
Updated On: Jun 16, 2026
- \(5\) sq. units
- \(\dfrac13\) sq. unit
- \(\dfrac15\) sq. unit
- \(3\) sq. units
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
Area under a piecewise-defined curve is obtained by integrating each piece separately over its interval.
Step 1: Split the area into two parts. \[ A = \int_0^1 x^2\,dx + \int_1^4 \sqrt{x}\,dx \]
Step 2: Evaluate the first integral. \[\begin{aligned} \int_0^1 x^2\,dx &= \left[\frac{x^3}{3}\right]_0^1 \\ &= \frac13 \end{aligned}\]
Step 3: Evaluate the second integral. \[\begin{aligned} \int_1^4 \sqrt{x}\,dx &= \int_1^4 x^{1/2}\,dx \\ &= \left[\frac{2}{3}x^{3/2}\right]_1^4 \\ &= \frac23(8-1) \\ &= \frac{14}{3} \end{aligned}\]
Step 4: Find the total area. \[\begin{aligned} A &= \frac13+\frac{14}{3} \\ &= \frac{15}{3} \\ &= 5 \end{aligned}\] \[\begin{aligned} \boxed{5} \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.
Step 1: Split the area into two parts. \[ A = \int_0^1 x^2\,dx + \int_1^4 \sqrt{x}\,dx \]
Step 2: Evaluate the first integral. \[\begin{aligned} \int_0^1 x^2\,dx &= \left[\frac{x^3}{3}\right]_0^1 \\ &= \frac13 \end{aligned}\]
Step 3: Evaluate the second integral. \[\begin{aligned} \int_1^4 \sqrt{x}\,dx &= \int_1^4 x^{1/2}\,dx \\ &= \left[\frac{2}{3}x^{3/2}\right]_1^4 \\ &= \frac23(8-1) \\ &= \frac{14}{3} \end{aligned}\]
Step 4: Find the total area. \[\begin{aligned} A &= \frac13+\frac{14}{3} \\ &= \frac{15}{3} \\ &= 5 \end{aligned}\] \[\begin{aligned} \boxed{5} \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.
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