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Area Under the Curves question

2019 · 9 Apr · Shift 1 · Q38
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  5. /2019 · 9 Apr · Shift 1 · Q38

Area Under the Curves question

2019 · 9 Apr · Shift 1 · Q38

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area (in sq. units) of the region A = {(x, y) : x2 ≤\le≤ y ≤\le≤ x + 2} is
  1. A
    316{{31 \over 6}}631​
  2. B
    103{{10 \over 3}}310​
  3. C
    136{{13 \over 6}}613​
  4. D
    92{{9 \over 2}}29​
View written solutionFree

Correct answer: D

  1. Interpret the region

The region is

A={(x,y):x2≤y≤x+2}.A=\{(x,y): x^2 \le y \le x+2\}.A={(x,y):x2≤y≤x+2}.

So for a given xxx, the curve y=x2y=x^2y=x2 is the lower boundary and the line y=x+2y=x+2y=x+2 is the upper boundary.

For the region to exist, we need

x2≤x+2.x^2 \le x+2.x2≤x+2.
  1. Find the points of intersection

Solve

x2=x+2.x^2=x+2.x2=x+2.

This gives

x2−x−2=0x^2-x-2=0x2−x−2=0 (x−2)(x+1)=0.(x-2)(x+1)=0.(x−2)(x+1)=0.

Hence,

x=−1,  2.x=-1,\;2.x=−1,2.

So the curves intersect at x=−1x=-1x=−1 and x=2x=2x=2.

  1. Set up the area integral

Area between the curves is

∫−12[(x+2)−x2] dx.\int_{-1}^{2} \big[(x+2)-x^2\big] \, dx.∫−12​[(x+2)−x2]dx.
  1. Evaluate the integral
∫−12(x+2−x2) dx=[x22+2x−x33]−12.\int_{-1}^{2} (x+2-x^2)\,dx =\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}.∫−12​(x+2−x2)dx=[2x2​+2x−3x3​]−12​.

At x=2x=2x=2:

222+2(2)−233=2+4−83=6−83=103.\frac{2^2}{2}+2(2)-\frac{2^3}{3} =2+4-\frac{8}{3} =6-\frac{8}{3} =\frac{10}{3}.222​+2(2)−323​=2+4−38​=6−38​=310​.

At x=−1x=-1x=−1:

(−1)22+2(−1)−(−1)33=12−2+13=3+2−126=−76.\frac{(-1)^2}{2}+2(-1)-\frac{(-1)^3}{3} =\frac12-2+\frac13 =\frac{3+2-12}{6} =-\frac76.2(−1)2​+2(−1)−3(−1)3​=21​−2+31​=63+2−12​=−67​.

Therefore,

Area=103−(−76)=206+76=276=92.\text{Area}=\frac{10}{3}-\left(-\frac{7}{6}\right) =\frac{20}{6}+\frac{7}{6} =\frac{27}{6} =\frac92.Area=310​−(−67​)=620​+67​=627​=29​.
  1. Compare with options
92\frac9229​

corresponds to Option D.

  1. Comparison with stored answer

Stored correct answer: D.

This matches our derived answer.

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