The area (in sq. units) of the region described by A= (x, y) | . y \ge x2 - 5x + 4, x + y \ge 1, y \le 0 is :

Area Under the Curves PYQ — JEE Main 2016 · 9 Apr · Shift 1 · Q39 with Solution

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Correct answer: B

Interpret the region

We are given $A=\{(x,y)\mid y\ge x^2-5x+4,\; x+y\ge 1,\; y\le 0\}.$

Rewrite the line condition: $x+y\ge 1 \implies y\ge 1-x.$

So the region satisfies:

$y\ge x^2-5x+4$ (above the parabola)
$y\ge 1-x$ (above the line)
$y\le 0$ (below the $x$ -axis)

Hence, for each $x$ , the lower boundary is the larger of $x^2-5x+4 \quad \text{and} \quad 1-x,$ and the upper boundary is $y=0$ .

So the region exists where $\max(x^2-5x+4,\,1-x) \le 0.$

Find where the parabola and line intersect

Set $x^2-5x+4=1-x.$ Then $x^2-4x+3=0$ $(x-1)(x-3)=0.$ So they intersect at $x=1,\;3.$

Corresponding $y$ -values from $y=1-x$ are:

at $x=1$ , $y=0$
at $x=3$ , $y=-2$

Thus intersection points are $(1,0),\;(3,-2).$

Find where each curve is below the $x$ -axis

Since $y\le 0$ , we need the lower boundary also to be at most $0$ .

For the line:

$1-x\le 0 \implies x\ge 1.$

For the parabola:

$x^2-5x+4\le 0$ $(x-1)(x-4)\le 0 \implies 1\le x\le 4.$

Therefore the possible $x$ -range is $1\le x\le 4.$

Determine which curve is higher (i.e. larger)

Compare $f(x)=x^2-5x+4,\qquad g(x)=1-x.$

Compute $f(x)-g(x)=x^2-5x+4-(1-x)=x^2-4x+3=(x-1)(x-3).$

So:

for $1<x<3$ , $(x-1)(x-3)<0$ , hence $f(x)<g(x)$ , so the larger curve is $g(x)=1-x$ .
for $3<x<4$ , $(x-1)(x-3)>0$ , hence $f(x)>g(x)$ , so the larger curve is $f(x)=x^2-5x+4$ .

Thus the area is split into two parts: $\text{Area}=\int_1^3 \big(0-(1-x)\big)\,dx + \int_3^4 \big(0-(x^2-5x+4)\big)\,dx.$

That is, $\text{Area}=\int_1^3 (x-1)\,dx + \int_3^4 (-x^2+5x-4)\,dx.$

Evaluate the first integral

$\int_1^3 (x-1)\,dx = \left[\frac{x^2}{2}-x\right]_1^3.$

At $x=3$ : $\frac{9}{2}-3=\frac{3}{2}.$

At $x=1$ : $\frac{1}{2}-1=-\frac{1}{2}.$

So, $\int_1^3 (x-1)\,dx=\frac{3}{2}-\left(-\frac{1}{2}\right)=2.$

Evaluate the second integral

$\int_3^4 (-x^2+5x-4)\,dx=\left[-\frac{x^3}{3}+\frac{5x^2}{2}-4x\right]_3^4.$

At $x=4$ : $-\frac{64}{3}+40-16=24-\frac{64}{3}=\frac{8}{3}.$

At $x=3$ : $-9+\frac{45}{2}-12=\frac{3}{2}.$

So, $\int_3^4 (-x^2+5x-4)\,dx=\frac{8}{3}-\frac{3}{2}=\frac{16-9}{6}=\frac{7}{6}.$

Total area

$\text{Area}=2+\frac{7}{6}=\frac{12}{6}+\frac{7}{6}=\frac{19}{6}.$

So the required area is $\boxed{\frac{19}{6}}.$

Compare with stored answer

Stored correct answer: B

Our derived answer is $\frac{19}{6},$ which corresponds to Option B.

Therefore, the stored answer is correct.

Area Under the Curves question

Area Under the Curves question

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For the parabola:

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