Application of IntegralsmediumFree

Application of Integrals: Find Area Region Bounded Curves Lies Right Line

JEE Maths question with a full step-by-step solution.

Question
Find the area of the region bounded by the curves y=x2y = x^2, y=∣2−x2∣y = \left|2-x^2\right| and y=2y = 2, which lies to the right of the line x=1x = 1.
Solution
Answer: 1.01 (± 0.01)
Question attachment Step 1:
∣2−x2∣={2−x2,∣x∣≤2,x2−2,∣x∣>2.\left|2-x^2\right| = \begin{cases} 2-x^2 , & \left|x\right|\le\sqrt2 ,\\ x^2-2 , & \left|x\right|>\sqrt2 .\end{cases}
 So the description of the region changes at x=2x = \sqrt2. At x=1x = 1 the two curves meet, x2=1=∣2−x2∣x^2 = 1 = \left|2-x^2\right|, so the region closes on the left. Step 2: For x>1x>1,
y=x2 meets y=2 at x=2;y=x2−2 meets y=2 at x=2y = x^2 \ \text{meets}\ y = 2 \ \text{at}\ x = \sqrt2 ;\qquad y = x^2-2 \ \text{meets}\ y = 2 \ \text{at}\ x = 2
 So the region runs from x=1x = 1 to x=2x = 2. Step 3: Case-I: 1≤x≤21\le x\le\sqrt2. Here y=x2y = x^2 rises from 11 to 22 and ∣2−x2∣=2−x2\left|2-x^2\right| = 2-x^2 falls from 11 to 00, and x2≥2−x2x^2 \ge 2-x^2 for x≥1x\ge1. The line y=2y = 2 is above both. So the strip is bounded above by x2x^2 and below by 2−x22-x^2:
A1=∫12[x2−(2−x2)]dx=∫12(2x2−2)dxA_1 = \int_1^{\sqrt2}\left[x^2-\left(2-x^2\right)\right]dx = \int_1^{\sqrt2}\left(2x^2-2\right)dx
 Step 4: Case-II: 2≤x≤2\sqrt2\le x\le2. Now x2≥2x^2 \ge 2, so the parabola y=x2y = x^2 has left the strip and y=2y = 2 is the ceiling, while ∣2−x2∣=x2−2\left|2-x^2\right| = x^2-2 is the floor, x2−2≤2x^2-2 \le 2 holding for x≤2x\le2:
A2=∫22[2−(x2−2)]dx=∫22(4−x2)dxA_2 = \int_{\sqrt2}^{2}\left[2-\left(x^2-2\right)\right]dx = \int_{\sqrt2}^{2}\left(4-x^2\right)dx
 Step 5:
A1=[2x33−2x]12=(423−22)−(23−2)=−223+43=4−223A_1 = \left[\frac{2x^3}{3}-2x\right]_1^{\sqrt2} = \left(\frac{4\sqrt2}{3}-2\sqrt2\right)-\left(\frac23-2\right) = -\frac{2\sqrt2}{3}+\frac43 = \frac{4-2\sqrt2}{3}
A2=[4x−x33]22=(8−83)−(42−223)=163−1023=16−1023A_2 = \left[4x-\frac{x^3}{3}\right]_{\sqrt2}^{2} = \left(8-\frac83\right)-\left(4\sqrt2-\frac{2\sqrt2}{3}\right) = \frac{16}{3}-\frac{10\sqrt2}{3} = \frac{16-10\sqrt2}{3}
 Step 6: On adding,
A=4−22+16−1023=20−1223=203−42A = \frac{4-2\sqrt2+16-10\sqrt2}{3} = \frac{20-12\sqrt2}{3} = \frac{20}{3}-4\sqrt2
≈6.66667−5.65685=1.00981  ⟶  1.01\approx 6.66667-5.65685 = 1.00981 \;\longrightarrow\; 1.01
 Answer: 1.011.01 (exactly 203−42\dfrac{20}{3}-4\sqrt2)
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