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Maths Question 19 – JEE-MAIN 2026

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The area of the region (x,y):x2−8x≤y≤−x is:

The region is bounded by two curves, y1=x2−8x and y2=−x.

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

The region is bounded by the parabola y=x2−8x and the line y=−x. To find the points of intersection, we set the equations equal to each other:

x2−8x=−x

Rearranging the terms, we get:

x2−7x=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=x2−8x, at x=1, y(1)=12−8(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=x2−8x 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=∫07(−x−(x2−8x))dx

Simplify the integrand:

A=∫07(−x−x2+8x)dx=∫07(7x−x2)dx

Now, perform the integration:

A=[7x22−x33]07

Evaluate the definite integral at the limits:

A=(7(7)22−(7)33)−(7(0)22−(0)33)
A=732−733=343(12−13)
A=343(3−26)=343(16)=3436

The area of the region is 3436.

💡 Teacher's Secret Hint

Remember the fundamental theorem of calculus for evaluating definite integrals.

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