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Area Under the Curves question

2021 · 25 Jul · Shift 1 · Q31
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  5. /2021 · 25 Jul · Shift 1 · Q31

Area Under the Curves question

2021 · 25 Jul · Shift 1 · Q31

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area (in sq. units) of the region, given by the set {(x,y)∈R×R∣x≥0,2x2≤y≤4−2x}\{ (x,y) \in R \times R|x \ge 0,2{x^2} \le y \le 4 - 2x\}{(x,y)∈R×R∣x≥0,2x2≤y≤4−2x} is :
  1. A
    83{8 \over 3}38​
  2. B
    173{{17} \over 3}317​
  3. C
    133{{13} \over 3}313​
  4. D
    73{7 \over 3}37​
View written solutionFree

Correct answer: D

  1. Interpret the region

The region is given by

{(x,y)∈R×R∣x≥0,  2x2≤y≤4−2x}.\{(x,y)\in \mathbb R\times \mathbb R\mid x\ge 0,\; 2x^2 \le y \le 4-2x\}.{(x,y)∈R×R∣x≥0,2x2≤y≤4−2x}.

So for each admissible xxx, the curve y=2x2y=2x^2y=2x2 is the lower boundary and the line y=4−2xy=4-2xy=4−2x is the upper boundary.

Hence the required area is the area enclosed between these two curves for those xxx where 2x2≤4−2x.2x^2 \le 4-2x.2x2≤4−2x.

  1. Find the points of intersection

Set the two curves equal:

2x2=4−2x.2x^2 = 4-2x.2x2=4−2x.

Divide by 222:

x2=2−x.x^2 = 2-x.x2=2−x.

Rearrange:

x2+x−2=0.x^2+x-2=0.x2+x−2=0.

Factorize:

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

So the intersection points occur at x=−2,  1.x=-2,\; 1.x=−2,1.

But the region also has the condition x≥0x\ge 0x≥0, so only 0≤x≤10\le x\le 10≤x≤1 is relevant.

  1. Set up the area integral

Area between curves =

∫01[(upper curve)−(lower curve)]dx.\int_{0}^{1} \big[(\text{upper curve})-(\text{lower curve})\big]dx.∫01​[(upper curve)−(lower curve)]dx.

Thus,

A=∫01[(4−2x)−2x2]dx.A=\int_0^1 \big[(4-2x)-2x^2\big]dx.A=∫01​[(4−2x)−2x2]dx.
  1. Evaluate the integral
A=∫01(4−2x−2x2) dx.A=\int_0^1 (4-2x-2x^2)\,dx.A=∫01​(4−2x−2x2)dx.

Antiderivative:

∫(4−2x−2x2)dx=4x−x2−2x33.\int (4-2x-2x^2)dx = 4x-x^2-\frac{2x^3}{3}.∫(4−2x−2x2)dx=4x−x2−32x3​.

Now apply the limits 000 to 111:

A=[4x−x2−2x33]01.A=\left[4x-x^2-\frac{2x^3}{3}\right]_0^1.A=[4x−x2−32x3​]01​.

At x=1x=1x=1:

4(1)−12−2(1)33=4−1−23=3−23=73.4(1)-1^2-\frac{2(1)^3}{3}=4-1-\frac{2}{3}=3-\frac{2}{3}=\frac{7}{3}.4(1)−12−32(1)3​=4−1−32​=3−32​=37​.

At x=0x=0x=0:

0.0.0.

Therefore,

A=73.A=\frac{7}{3}.A=37​.
  1. Match with the options
73\frac{7}{3}37​

corresponds to Option D.

  1. Compare with stored correct answer

Stored correct answer: D

Our derived answer: D

So they agree.

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