Let f : [-3, 1] \to \mathbbR be given as f(x) = \ \beginarrayll \min\(x + 6), x^2\ & -3 \le x \le 0 \max sqrtx, x^2\ & 0 \le x \le 1 \endarray . If the area bounded by y = f(x) and x -axis is A , then the value of 6A is equal to \underline\hspace2cm .

Area Under the Curves PYQ — JEE Main 2021 · 17 Mar · Shift 2 · Q42 with Solution

View written solutionFree

Correct answer: 41

Understand the piecewise function

Given

f(x)= \begin{cases} \min\{x+6,\,x^2\}, & -3\le x\le 0 \\ \max\{\sqrt{x},\,x^2\}, & 0\le x\le 1 \end{cases}

We need the area bounded by $y=f(x)$ and the $x$ -axis on $[-3,1]$ .

Since on both intervals $f(x)\ge 0$ , the required area is simply

A=\int_{-3}^{1} f(x)\,dx.

For $-3\le x\le 0$ : find $\min\{x+6,x^2\}$

We compare $x+6$ and $x^2$ :

x^2=x+6

x^2-x-6=0

(x-3)(x+2)=0

So intersection points are $x=3,-2$ . In the interval $[-3,0]$ , only $x=-2$ is relevant.

Now check which is smaller:

At $x=-3$ : $x^2=9$ , $x+6=3$ so minimum is $x+6$ .
At $x=-1$ : $x^2=1$ , $x+6=5$ so minimum is $x^2$ .

Hence,

f(x)= \begin{cases} x+6, & -3\le x\le -2 \\ x^2, & -2\le x\le 0 \end{cases}

So area from this part is

A_1=\int_{-3}^{-2} (x+6)\,dx+\int_{-2}^{0} x^2\,dx.

Compute:

\int_{-3}^{-2} (x+6)\,dx=\left[\frac{x^2}{2}+6x\right]_{-3}^{-2} =\left(2-12\right)-\left(\frac{9}{2}-18\right) =-10+\frac{27}{2}=\frac{7}{2}.

And

\int_{-2}^{0} x^2\,dx=\left[\frac{x^3}{3}\right]_{-2}^{0}=0-\left(-\frac{8}{3}\right)=\frac{8}{3}.

Thus

A_1=\frac{7}{2}+\frac{8}{3}=\frac{21+16}{6}=\frac{37}{6}.

For $0\le x\le 1$ : find $\max\{\sqrt{x},x^2\}$

Compare $\sqrt{x}$ and $x^2$ :

\sqrt{x}=x^2

For $x\ge 0$ , squaring gives

x=x^4

x(x^3-1)=0

So $x=0$ or $x=1$ .

Now for $0<x<1$ ,

\sqrt{x}>x^2

(for example at $x=\tfrac14$ , $\sqrt{x}=\tfrac12$ and $x^2=\tfrac1{16}$ ).

Therefore,

f(x)=\sqrt{x} \quad \text{for } 0\le x\le 1.

So area from this part is

A_2=\int_0^1 \sqrt{x}\,dx=\int_0^1 x^{1/2}\,dx =\left[\frac{2}{3}x^{3/2}\right]_0^1=\frac{2}{3}.

Total area

A=A_1+A_2=\frac{37}{6}+\frac{2}{3}=\frac{37}{6}+\frac{4}{6}=\frac{41}{6}.

Therefore,

6A=6\cdot \frac{41}{6}=41.

Compare with stored answer

Derived answer: $41$

Stored correct answer: $41$

They match.

More from Area Under the Curves

View written solutionFree

Correct answer: 41

Understand the piecewise function

Given

f(x)= \begin{cases} \min\{x+6,\,x^2\}, & -3\le x\le 0 \\ \max\{\sqrt{x},\,x^2\}, & 0\le x\le 1 \end{cases}

We need the area bounded by $y=f(x)$ and the $x$ -axis on $[-3,1]$ .

Since on both intervals $f(x)\ge 0$ , the required area is simply

A=\int_{-3}^{1} f(x)\,dx.

For $-3\le x\le 0$ : find $\min\{x+6,x^2\}$

We compare $x+6$ and $x^2$ :

x^2=x+6

x^2-x-6=0

(x-3)(x+2)=0

So intersection points are $x=3,-2$ . In the interval $[-3,0]$ , only $x=-2$ is relevant.

Now check which is smaller:

At $x=-3$ : $x^2=9$ , $x+6=3$ so minimum is $x+6$ .
At $x=-1$ : $x^2=1$ , $x+6=5$ so minimum is $x^2$ .

Hence,

f(x)= \begin{cases} x+6, & -3\le x\le -2 \\ x^2, & -2\le x\le 0 \end{cases}

So area from this part is

A_1=\int_{-3}^{-2} (x+6)\,dx+\int_{-2}^{0} x^2\,dx.

Compute:

\int_{-3}^{-2} (x+6)\,dx=\left[\frac{x^2}{2}+6x\right]_{-3}^{-2} =\left(2-12\right)-\left(\frac{9}{2}-18\right) =-10+\frac{27}{2}=\frac{7}{2}.

And

\int_{-2}^{0} x^2\,dx=\left[\frac{x^3}{3}\right]_{-2}^{0}=0-\left(-\frac{8}{3}\right)=\frac{8}{3}.

Thus

A_1=\frac{7}{2}+\frac{8}{3}=\frac{21+16}{6}=\frac{37}{6}.

For $0\le x\le 1$ : find $\max\{\sqrt{x},x^2\}$

Compare $\sqrt{x}$ and $x^2$ :

\sqrt{x}=x^2

For $x\ge 0$ , squaring gives

x=x^4

x(x^3-1)=0

So $x=0$ or $x=1$ .

Now for $0<x<1$ ,

\sqrt{x}>x^2

(for example at $x=\tfrac14$ , $\sqrt{x}=\tfrac12$ and $x^2=\tfrac1{16}$ ).

Therefore,

f(x)=\sqrt{x} \quad \text{for } 0\le x\le 1.

So area from this part is

A_2=\int_0^1 \sqrt{x}\,dx=\int_0^1 x^{1/2}\,dx =\left[\frac{2}{3}x^{3/2}\right]_0^1=\frac{2}{3}.

Total area

A=A_1+A_2=\frac{37}{6}+\frac{2}{3}=\frac{37}{6}+\frac{4}{6}=\frac{41}{6}.

Therefore,

6A=6\cdot \frac{41}{6}=41.

Compare with stored answer

Derived answer: $41$

Stored correct answer: $41$

They match.

Area Under the Curves question

Area Under the Curves question

More from Area Under the Curves

Area Under the Curves question

Area Under the Curves question

More from Area Under the Curves