The area (in sq. units) of the region (x, y) : 0 \le y \le x2 + 1, 0 \le y \le x + 1, 1 \over 2 \le x \le 2 is :

Area Under the Curves PYQ — JEE Main 2020 · 3 Sep · Shift 1 · Q32 with Solution

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Correct answer: B

The region is given by $0\le y\le x^2+1,\qquad 0\le y\le x+1,\qquad \frac12\le x\le 2.$

Since $y$ must satisfy both upper bounds, for each fixed $x$ we have $0\le y\le \min(x^2+1,\,x+1).$

So the required area is $\int_{1/2}^{2} \min(x^2+1,\,x+1)\,dx.$
Find where the two curves intersect: $x^2+1=x+1$ $x^2-x=0$ $x(x-1)=0$ Hence, intersection points occur at $x=0,1$ .

In the interval $\left[\frac12,2\right]$ , the relevant switching point is $x=1$ .
Compare the two functions on the subintervals:

$x^2+1-(x+1)=x^2-x=x(x-1).$
- For $\frac12\le x<1$ , we have $x>0$ and $x-1<0$ , so $x(x-1)<0$ . Thus, $x^2+1<x+1,$ so the lower upper-bound is $x^2+1$ .
- For $1<x\le 2$ , we have $x(x-1)>0$ , so $x^2+1>x+1,$ and the lower upper-bound is $x+1$ .
Therefore the area is $A=\int_{1/2}^{1}(x^2+1)\,dx+\int_{1}^{2}(x+1)\,dx.$
Evaluate the first integral: $\int (x^2+1)dx=\frac{x^3}{3}+x.$ So,
$\left(\frac{x^3}{3}+x\right)_{1/2}^{1}.$$ At $x=1$: $$\frac{1}{3}+1=\frac{4}{3}.$$ At $x=\frac12$: $$\frac{(1/2)^3}{3}+\frac12=\frac{1}{24}+\frac{12}{24}=\frac{13}{24}.$$ Hence, $$\int_{1/2}^{1}(x^2+1)dx=\frac{4}{3}-\frac{13}{24}= rac{32-13}{24}=\frac{19}{24}.$$$
Evaluate the second integral: $\int (x+1)dx=\frac{x^2}{2}+x.$ So,
$\left(\frac{x^2}{2}+x\right)_{1}^{2}.$$ At $x=2$: $$\frac{4}{2}+2=4.$$ At $x=1$: $$\frac{1}{2}+1=\frac{3}{2}.$$ Therefore, $$\int_{1}^{2}(x+1)dx=4-\frac32=\frac52.$$$
Total area: $A=\frac{19}{24}+\frac52= rac{19}{24}+\frac{60}{24}=\frac{79}{24}.$
Hence the correct option is $\boxed{\frac{79}{24}}$ which is Option B.

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View written solutionFree

Correct answer: B

The region is given by $0\le y\le x^2+1,\qquad 0\le y\le x+1,\qquad \frac12\le x\le 2.$

Since $y$ must satisfy both upper bounds, for each fixed $x$ we have $0\le y\le \min(x^2+1,\,x+1).$

So the required area is $\int_{1/2}^{2} \min(x^2+1,\,x+1)\,dx.$
Find where the two curves intersect: $x^2+1=x+1$ $x^2-x=0$ $x(x-1)=0$ Hence, intersection points occur at $x=0,1$ .

In the interval $\left[\frac12,2\right]$ , the relevant switching point is $x=1$ .
Compare the two functions on the subintervals:

$x^2+1-(x+1)=x^2-x=x(x-1).$
- For $\frac12\le x<1$ , we have $x>0$ and $x-1<0$ , so $x(x-1)<0$ . Thus, $x^2+1<x+1,$ so the lower upper-bound is $x^2+1$ .
- For $1<x\le 2$ , we have $x(x-1)>0$ , so $x^2+1>x+1,$ and the lower upper-bound is $x+1$ .
Therefore the area is $A=\int_{1/2}^{1}(x^2+1)\,dx+\int_{1}^{2}(x+1)\,dx.$
Evaluate the first integral: $\int (x^2+1)dx=\frac{x^3}{3}+x.$ So,
$\left(\frac{x^3}{3}+x\right)_{1/2}^{1}.$$ At $x=1$: $$\frac{1}{3}+1=\frac{4}{3}.$$ At $x=\frac12$: $$\frac{(1/2)^3}{3}+\frac12=\frac{1}{24}+\frac{12}{24}=\frac{13}{24}.$$ Hence, $$\int_{1/2}^{1}(x^2+1)dx=\frac{4}{3}-\frac{13}{24}= rac{32-13}{24}=\frac{19}{24}.$$$
Evaluate the second integral: $\int (x+1)dx=\frac{x^2}{2}+x.$ So,
$\left(\frac{x^2}{2}+x\right)_{1}^{2}.$$ At $x=2$: $$\frac{4}{2}+2=4.$$ At $x=1$: $$\frac{1}{2}+1=\frac{3}{2}.$$ Therefore, $$\int_{1}^{2}(x+1)dx=4-\frac32=\frac52.$$$
Total area: $A=\frac{19}{24}+\frac52= rac{19}{24}+\frac{60}{24}=\frac{79}{24}.$
Hence the correct option is $\boxed{\frac{79}{24}}$ which is Option B.

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