The area of the region (x, y): x^2 - 8x <= y <= -x is:

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q19 – JEE-MAIN 2026

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Home Maths Limits and DerivativesQuestion #267

The area of the region

(x, y) : x^{2} - 8 x \leq y \leq - x

is:

Clue 1 of 3: Identify the Bounding Curves

The region is bounded by two curves, $y_{1} = x^{2} - 8 x$ and $y_{2} = - x$ .

🧠 Full Solution Path

● Video Walkthrough

Step 1: Find the intersection points of the curves✦ Active

The region is bounded by the parabola $y = x^{2} - 8 x$ and the line $y = - x$ . To find the points of intersection, we set the equations equal to each other:

x^{2} - 8 x = - x

Rearranging the terms, we get:

x^{2} - 7 x = 0

Factoring out $x$ gives:

x (x - 7) = 0

Thus, the intersection points are at $x = 0$ and $x = 7$ . These will serve as the limits of integration.

Step 2: Determine the upper and lower curves○ Expand

To set up the integral correctly, we need to determine which function is the upper curve and which is the lower curve in the interval $[0, 7]$ . We can test a point within this interval, for example, $x = 1$ :

For the parabola $y = x^{2} - 8 x$ , at $x = 1$ , $y (1) = 1^{2} - 8 (1) = 1 - 8 = - 7$ .

For the line $y = - x$ , at $x = 1$ , $y (1) = - 1$ .

Since $- 1 > - 7$ , the line $y = - x$ is the upper curve and the parabola $y = x^{2} - 8 x$ is the lower curve in the interval $[0, 7]$ .

💡 Teacher's Secret Hint

Always verify the relative positions of the curves within the integration interval to avoid sign errors.

Step 3: Calculate the area using definite integration○ Expand

The area $A$ between the two curves is given by the integral of the upper curve minus the lower curve from $x = 0$ to $x = 7$ :

A = \int_{0}^{7} (- x - (x^{2} - 8 x)) d x

Simplify the integrand:

A = \int_{0}^{7} (- x - x^{2} + 8 x) d x = \int_{0}^{7} (7 x - x^{2}) d x

Now, perform the integration:

A = {[\frac{7 x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{7}

Evaluate the definite integral at the limits:

A = (\frac{7 (7)^{2}}{2} - \frac{(7)^{3}}{3}) - (\frac{7 (0)^{2}}{2} - \frac{(0)^{3}}{3})

A = \frac{7^{3}}{2} - \frac{7^{3}}{3} = 343 (\frac{1}{2} - \frac{1}{3})

A = 343 (\frac{3 - 2}{6}) = 343 (\frac{1}{6}) = \frac{343}{6}

The area of the region is $\frac{343}{6}$ .

💡 Teacher's Secret Hint

Remember the fundamental theorem of calculus for evaluating definite integrals.

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Maths Question 19 – JEE-MAIN 2026

Clue 1 of 3: Identify the Bounding Curves

🧠 Full Solution Path