Let A be the area bounded by the curve y=x|x-3| , the x -axis and the ordinates x=-1 and x=2 . Then 12 A is equal to \underline\hspace2cm .

Area Under the Curves PYQ — JEE Main 2023 · 1 Feb · Shift 1 · Q41 with Solution

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Correct answer: 62

Given curve

$y=x|x-3|$

We need the area bounded by this curve, the $x$ -axis, and the vertical lines $x=-1$ and $x=2$ .
Remove the modulus

On the interval $[-1,2]$ , we have $x<3$ , so

$|x-3|=3-x$

Hence the curve becomes

$y=x(3-x)=3x-x^2$
Check where the curve is above/below the $x$ -axis

$y=x(3-x)$

Its zeros are at $x=0$ and $x=3$ .

On $[-1,0]$ , $x<0$ and $3-x>0$ , so $y<0$ . On $[0,2]$ , $x>0$ and $3-x>0$ , so $y>0$ .

Therefore area is

$A= ext{area on }[-1,0]+ ext{area on }[0,2]$
$\int_{-1}^{0}(3x-x^2)\,dx+\int_{0}^{2}(3x-x^2)\,dx$$$
Evaluate the integrals

Antiderivative:

$\int (3x-x^2)\,dx=\frac{3x^2}{2}-\frac{x^3}{3}$

First part:
$\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_{-1}^{0}$$ At $x=0$: $$0$$ At $x=-1$: $$\frac{3(1)}{2}-\frac{(-1)^3}{3}=\frac{3}{2}+\frac{1}{3}=\frac{11}{6}$$ So $$\int_{-1}^{0}(3x-x^2)\,dx=0-\frac{11}{6}=-\frac{11}{6}$$ Therefore corresponding area is $$\frac{11}{6}$$ **Second part:** $$\int_{0}^{2}(3x-x^2)\,dx= \left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_{0}^{2}$$ $$=\left(\frac{3\cdot 4}{2}-\frac{8}{3}\right)-0=6-\frac{8}{3}=\frac{10}{3}$$$
Total area

$A=\frac{11}{6}+\frac{10}{3}=\frac{11}{6}+\frac{20}{6}=\frac{31}{6}$
Compute $12A$

$12A=12\cdot \frac{31}{6}=2\cdot 31=62$
Comparison with stored answer

Our derived answer is $62$ , which matches the stored correct answer.

View written solutionFree

Correct answer: 62

Given curve

$y=x|x-3|$

We need the area bounded by this curve, the $x$ -axis, and the vertical lines $x=-1$ and $x=2$ .
Remove the modulus

On the interval $[-1,2]$ , we have $x<3$ , so

$|x-3|=3-x$

Hence the curve becomes

$y=x(3-x)=3x-x^2$
Check where the curve is above/below the $x$ -axis

$y=x(3-x)$

Its zeros are at $x=0$ and $x=3$ .

On $[-1,0]$ , $x<0$ and $3-x>0$ , so $y<0$ . On $[0,2]$ , $x>0$ and $3-x>0$ , so $y>0$ .

Therefore area is

$A= ext{area on }[-1,0]+ ext{area on }[0,2]$
$\int_{-1}^{0}(3x-x^2)\,dx+\int_{0}^{2}(3x-x^2)\,dx$$$
Evaluate the integrals

Antiderivative:

$\int (3x-x^2)\,dx=\frac{3x^2}{2}-\frac{x^3}{3}$

First part:
$\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_{-1}^{0}$$ At $x=0$: $$0$$ At $x=-1$: $$\frac{3(1)}{2}-\frac{(-1)^3}{3}=\frac{3}{2}+\frac{1}{3}=\frac{11}{6}$$ So $$\int_{-1}^{0}(3x-x^2)\,dx=0-\frac{11}{6}=-\frac{11}{6}$$ Therefore corresponding area is $$\frac{11}{6}$$ **Second part:** $$\int_{0}^{2}(3x-x^2)\,dx= \left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_{0}^{2}$$ $$=\left(\frac{3\cdot 4}{2}-\frac{8}{3}\right)-0=6-\frac{8}{3}=\frac{10}{3}$$$
Total area

$A=\frac{11}{6}+\frac{10}{3}=\frac{11}{6}+\frac{20}{6}=\frac{31}{6}$
Compute $12A$

$12A=12\cdot \frac{31}{6}=2\cdot 31=62$
Comparison with stored answer

Our derived answer is $62$ , which matches the stored correct answer.