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

• A. \(5\) sq. units
• B. \(\dfrac13\) sq. unit
• C. \(\dfrac15\) sq. unit
• D. \(3\) sq. units

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.

Answer 1

The Correct Option is A

Step 1: Understand the region.
The curve changes its rule at $x=1$: it is $x^2$ on $[0,1]$ and $\sqrt{x}$ on $(1,4]$. The area under it from $x=0$ to $x=4$ is the sum of the two separate pieces.

Step 2: Split the area.
$A=\displaystyle\int_0^1 x^2\,dx+\int_1^4 \sqrt{x}\,dx$.

Step 3: First piece.
$\displaystyle\int_0^1 x^2\,dx=\left[\dfrac{x^3}{3}\right]_0^1=\dfrac{1}{3}$.

Step 4: Second piece.
$\displaystyle\int_1^4 x^{1/2}\,dx=\left[\dfrac{2}{3}x^{3/2}\right]_1^4=\dfrac{2}{3}\big(8-1\big)=\dfrac{14}{3}$.

Step 5: Add the two parts.
$A=\dfrac{1}{3}+\dfrac{14}{3}=\dfrac{15}{3}=5$.

Step 6: State the area.
The total area is $5$ square units. \[ \boxed{5\ \text{sq. units}} \]