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Area Under the Curves question

2019 · 9 Apr · Shift 2 · Q28
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  5. /2019 · 9 Apr · Shift 2 · Q28

Area Under the Curves question

2019 · 9 Apr · Shift 2 · Q28

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area (in sq. units) of the region A = {(x, y) : y22≤{{y{}^2} \over 2} \le2y2​≤ x ≤\le≤ y + 4} is :-
  1. A
    30
  2. B
    18
  3. C
    533{{53} \over 3}353​
  4. D
    16
View written solutionFree

Correct answer: B

  1. Interpret the region

The region is

A={(x,y):y22≤x≤y+4}.A=\{(x,y): \frac{y^2}{2} \le x \le y+4\}.A={(x,y):2y2​≤x≤y+4}.

So for a fixed yyy, the horizontal strip runs from the curve x=y22x=\frac{y^2}{2}x=2y2​ to the line x=y+4.x=y+4.x=y+4.

Hence the area is

Area=∫(right boundary−left boundary) dy.\text{Area}=\int (\text{right boundary} - \text{left boundary})\,dy.Area=∫(right boundary−left boundary)dy.
  1. Find the limits of integration

The curves intersect where

y22=y+4.\frac{y^2}{2}=y+4.2y2​=y+4.

Multiply by 222:

y2=2y+8y^2=2y+8y2=2y+8 y2−2y−8=0y^2-2y-8=0y2−2y−8=0 (y−4)(y+2)=0.(y-4)(y+2)=0.(y−4)(y+2)=0.

So the intersection points occur at y=−2andy=4.y=-2 \quad \text{and} \quad y=4.y=−2andy=4.

Thus the required area is

∫−24[(y+4)−y22]dy.\int_{-2}^{4} \left[(y+4)-\frac{y^2}{2}\right]dy.∫−24​[(y+4)−2y2​]dy.
  1. Evaluate the integral
Area=∫−24(y+4−y22)dy.\text{Area}=\int_{-2}^{4} \left(y+4-\frac{y^2}{2}\right)dy.Area=∫−24​(y+4−2y2​)dy.

Antiderivative:

∫(y+4−y22)dy=y22+4y−y36.\int \left(y+4-\frac{y^2}{2}\right)dy =\frac{y^2}{2}+4y-\frac{y^3}{6}.∫(y+4−2y2​)dy=2y2​+4y−6y3​.

Now substitute the limits:

Area=[y22+4y−y36]−24.\text{Area}=\left[\frac{y^2}{2}+4y-\frac{y^3}{6}\right]_{-2}^{4}.Area=[2y2​+4y−6y3​]−24​.

At y=4y=4y=4:

162+16−646=8+16−323=24−323=403.\frac{16}{2}+16-\frac{64}{6}=8+16-\frac{32}{3}=24-\frac{32}{3}=\frac{40}{3}.216​+16−664​=8+16−332​=24−332​=340​.

At y=−2y=-2y=−2:

42−8−(−8)6=2−8+43=−6+43=−143.\frac{4}{2}-8-\frac{(-8)}{6}=2-8+\frac{4}{3}=-6+\frac{4}{3}=-\frac{14}{3}.24​−8−6(−8)​=2−8+34​=−6+34​=−314​.

Therefore,

Area=403−(−143)=543=18.\text{Area}=\frac{40}{3}-\left(-\frac{14}{3}\right)=\frac{54}{3}=18.Area=340​−(−314​)=354​=18.
  1. Match with options

The area is 181818 which corresponds to Option B.

  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

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