The area of the plane region bounded by y=√(x), x∈[0,1], y=x^2, x∈[1,2] and y=-x^2+2x+4, x∈[0,2] is

B. 19/3. Step 1: The top boundary is y=-x^2+2x+4; the bottom is y=√(x) on [0,1] and y=x^2 on [1,2]; the y -axis closes the region on the left and the curves meet at (2,4). Step 2: Area under the top curve: _0^2(-x^2+2x+4) dx=-8/3+4+8=28/3. Step 3: Subtract _0^1√(x) dx=2/3 and _1^2x^2 dx=7/3: Area =28/3-2/3-7/3=19/3. Check: On [0,2] the top curve lies between 4 and 5, so the area under it must lie between 8 and 10; 28/3 9.33 does. Trap: Dropping the constant 4 of the top curve and ignoring the bottom curves gives _0^2(-x^2+2x) dx=4/3, option (D).

Question 5 of 8 · Application of Definite Integration

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The area of the plane region bounded by $y = x, x \in [0, 1]$ , $y = x^{2}, x \in [1, 2]$ and $y = - x^{2} + 2 x + 4, x \in [0, 2]$ is

Answer (Detailed Solution Below)View answer ▾

Answer: Option B — $\frac{19}{3}$

Detailed Solution

The top boundary is $y = - x^{2} + 2 x + 4$ ; the bottom is $y = x$ on $[0, 1]$ and $y = x^{2}$ on $[1, 2]$ ; the $y$ -axis closes the region on the left and the curves meet at $(2, 4)$ .

Area under the top curve: $\int_{0}^{2} (- x^{2} + 2 x + 4) d x = - \frac{8}{3} + 4 + 8 = \frac{28}{3}$ .

Subtract $\int_{0}^{1} x d x = \frac{2}{3}$ and $\int_{1}^{2} x^{2} d x = \frac{7}{3}$ : Area $= \frac{28}{3} - \frac{2}{3} - \frac{7}{3} = \frac{19}{3}$ .

✓ Check

[0, 2]

the top curve lies between

4

and

5

, so the area under it must lie between

8

and

10

;

\frac{28}{3} \approx 9.33

does.

⚠ Common trap

Dropping the constant

4

of the top curve and ignoring the bottom curves gives

\int_{0}^{2} (- x^{2} + 2 x) d x = \frac{4}{3}

, option (D).

Quick tip

Area $= \int_{0}^{2} (- x^{2} + 2 x + 4) d x - \int_{0}^{1} x d x - \int_{1}^{2} x^{2} d x = \frac{28}{3} - \frac{2}{3} - \frac{7}{3}$

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Application of Definite Integration — Preparation

This question tests your understanding of Application of Definite Integration, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

This topic uses definite integrals to find the area of regions in the plane. It covers the area under a curve and above the x-axis, the area between a curve and the y-axis, the area between two curves, and the areas of standard regions bounded by circles, parabolas, ellipses and straight lines. Always start with a rough sketch of the region and find where the curves intersect, since those points become the limits of integration. For the area between two curves, integrate (upper curve - lower curve) between their intersection points. If part of a curve lies below the x-axis, take the absolute value of that part separately rather than integrating straight across, or the positive and negative areas will cancel. Useful standard results include the area of the ellipse x^2/a^2 + y^2/b^2 = 1, which is pi ab. Practise the area under the curve questions below, each with a sketch and a step-by-step solution.

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