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Area Under the Curves question

2019 · 8 Apr · Shift 1 · Q24
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  5. /2019 · 8 Apr · Shift 1 · Q24

Area Under the Curves question

2019 · 8 Apr · Shift 1 · Q24

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1
The area (in sq. units) of the region A = { (x, y) ∈\in∈ R × R| 0 ≤\le≤ x ≤\le≤ 3, 0 ≤\le≤ y ≤\le≤ 4, y ≤\le≤ x2 + 3x} is :
  1. A
    596{{59} \over 6}659​
  2. B
    263{{26} \over 3}326​
  3. C
    8
  4. D
    536{{53} \over 6}653​
View written solutionFree

Correct answer: A

  1. Interpret the region

The set is

A={(x,y)∈R×R∣0≤x≤3,  0≤y≤4,  y≤x2+3x}.A=\{(x,y)\in \mathbb R\times \mathbb R\mid 0\le x\le 3,\; 0\le y\le 4,\; y\le x^2+3x\}.A={(x,y)∈R×R∣0≤x≤3,0≤y≤4,y≤x2+3x}.

So we are inside the rectangle 0≤x≤3,0≤y≤4,0\le x\le 3,\quad 0\le y\le 4,0≤x≤3,0≤y≤4, and also below the curve y=x2+3x.y=x^2+3x.y=x2+3x.

Hence, for each fixed x∈[0,3]x\in[0,3]x∈[0,3], the allowed values of yyy are 0≤y≤min⁡(4, x2+3x).0\le y\le \min(4,\,x^2+3x).0≤y≤min(4,x2+3x).

Therefore, area is

Area=∫03min⁡(4,x2+3x) dx.\text{Area}=\int_0^3 \min(4,x^2+3x)\,dx.Area=∫03​min(4,x2+3x)dx.
  1. Find where the curve meets the line y=4y=4y=4

We solve

x2+3x=4.x^2+3x=4.x2+3x=4.

So,

x2+3x−4=0x^2+3x-4=0x2+3x−4=0 (x+4)(x−1)=0.(x+4)(x-1)=0.(x+4)(x−1)=0.

Thus,

In the interval [0,3][0,3][0,3], the relevant point is x=1.x=1.x=1.

Now:

  • for 0≤x≤10\le x\le 10≤x≤1, we have x2+3x≤4x^2+3x\le 4x2+3x≤4,
  • for 1≤x≤31\le x\le 31≤x≤3, we have x2+3x≥4x^2+3x\ge 4x2+3x≥4.

So,

Area=∫01(x2+3x) dx+∫134 dx.\text{Area}=\int_0^1 (x^2+3x)\,dx + \int_1^3 4\,dx.Area=∫01​(x2+3x)dx+∫13​4dx.
  1. Evaluate the integrals

First integral:

∫01(x2+3x) dx=[x33+3x22]01=13+32.\int_0^1 (x^2+3x)\,dx =\left[\frac{x^3}{3}+\frac{3x^2}{2}\right]_0^1 =\frac{1}{3}+\frac{3}{2}.∫01​(x2+3x)dx=[3x3​+23x2​]01​=31​+23​.

Taking LCM 666,

13+32=26+96=116.\frac{1}{3}+\frac{3}{2}=\frac{2}{6}+\frac{9}{6}=\frac{11}{6}.31​+23​=62​+69​=611​.

Second integral:

∫134 dx=4(3−1)=8.\int_1^3 4\,dx=4(3-1)=8.∫13​4dx=4(3−1)=8.

Therefore,

\text{Area}=\frac{11}{6}+8= rac{11}{6}+\frac{48}{6}=\frac{59}{6}.
  1. Match with the options
596\frac{59}{6}659​

corresponds to Option A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.

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