Previous year questions: Application of Integrals

Q1 (CBSE Board 2026, 5 marks)

Apply integration to the region enclosed by the curve y equal to the absolute value of x minus six, the x-axis and the two lines x equal to four and x equal to eight.

Answer: The expression x minus six changes sign at six. Between four and six the absolute value is six minus x, and its integral is the area of a triangle of base two and height two, which is two. Between six and eight the absolute value is x minus six, and its integral is again two. Adding the two pieces gives four square units. The sketch confirms that the region is two congruent triangles sitting on the x-axis.

Q2 (CBSE Board 2026, 5 marks)

Apply integration to the region enclosed by the curve y equal to x times the absolute value of x, by the x-axis and by the vertical lines x equal to minus two and x equal to two.

Answer: For negative x the absolute value is minus x, so the curve is y equal to minus x squared and it lies below the axis. For positive x the curve is y equal to x squared and it lies above the axis. The two pieces have equal areas by symmetry, and each is the integral of x squared from zero to two, which is eight thirds. The total area is twice eight thirds, which is sixteen thirds square units. The signed integral would give zero, so splitting at the origin is essential.

Q3 (CBSE Board 2026, 5 marks)

Apply integration to obtain the area of the region enclosed between the parabola y equal to x squared and the horizontal line y equal to two.

Answer: The parabola and the line meet where x squared equals two, that is at x equal to plus or minus root two. The line is above the parabola between these points, so the area is the integral from minus root two to root two of two minus x squared. The integrand is even, so the integral is twice the integral from zero to root two of two minus x squared. The antiderivative is two x minus x cubed over three, which at root two gives two root two minus two root two over three, equal to four root two over three. Doubling gives eight root two over three square units.

Q4 (CBSE Board 2026, 5 marks)

Apply integration to the region enclosed between the ellipse with equation x squared over four plus y squared over three equal to one and the pair of lines x equal to minus two and x equal to two.

Answer: On the given interval the lines x equal to minus two and x equal to two are the ends of the ellipse itself, so the region is the whole ellipse. The upper boundary is y equal to root three times the square root of one minus x squared over four. By symmetry the area is twice the integral from zero to two of that expression. Substituting x equal to two sine theta makes the integral the standard quarter-ellipse area, and the result is pi times the product of the semiaxes, that is two root three pi square units.

Q5 (CBSE Board 2024, 5 marks)

Apply integration to the region enclosed between the circle x squared plus y squared equal to sixteen and the two lines x equal to minus two and x equal to two.

Answer: The strip between the two chords x equal to minus two and x equal to two contains both the upper and the lower half of the circle, and it is symmetric about both axes as well. Its area is therefore four times the area in the first quadrant, that is four times the integral from zero to two of the square root of sixteen minus x squared. The antiderivative is half x root of sixteen minus x squared plus eight inverse sine of x over four. At the upper limit two the bracket equals root twelve plus eight times pi by six, which is two root three plus four pi by three, and at the lower limit zero it equals zero. Multiplying that bracket by the factor four gives eight root three plus sixteen pi by three square units. The factor four is essential: a factor two would count only the upper half of the strip.

Q6 (CBSE Board 2024, 5 marks)

Apply integration to the area enclosed by the curve y equal to the square root of four minus x squared, the two lines x equal to minus one and x equal to one and the x-axis.

Answer: The upper semicircle of radius two is integrated between the two lines. The antiderivative is half x root of four minus x squared plus two inverse sine of x over two. Evaluating at one gives half root three plus two times pi by six, and at minus one the same expression is negative in both terms. The difference is root three plus two pi by three, so the area is root three plus two pi by three square units.

Q7 (CBSE Board 2024, 5 marks)

Apply integration to the region enclosed by the line x minus two y equal to four, the lines x equal to one and x equal to six, and the x-axis.

Answer: Solve the line for y: y equal to x minus four whole divided by two. The line crosses the x-axis at x equal to four, so the curve is below the axis from one to four and above it from four to six, and the area is computed in two pieces with the signs adjusted. The first piece integrates four minus x over two from one to four and the second integrates x minus four over two from four to six. The first gives nine quarters and the second gives one, so the total area is thirteen quarters square units.

Q8 (CBSE Board 2023, 5 marks)

Apply integration to the region enclosed between the parabola y squared equal to four a x and its latus rectum.

Answer: The latus rectum of y squared equal to four a x is the line x equal to a, and it cuts the parabola at y equal to plus or minus two a. Using horizontal strips, the right boundary is x equal to a and the left boundary is x equal to y squared over four a, with y from minus two a to two a. The integrand a minus y squared over four a is even, so the integral is twice the integral from zero to two a. The antiderivative is a y minus y cubed over twelve a, which at two a gives two a squared minus two a squared over three, equal to four a squared over three. Doubling gives eight a squared over three square units.

Q9 (CBSE Board 2023, 3 marks)

Find the area of the region defined by the inequalities x squared plus y squared less than or equal to one and x plus y greater than or equal to one, using integration.

Answer: The first inequality is the unit disc and the second is the half plane above the line x plus y equal to one. The line meets the circle at one comma zero and zero comma one, so the required region is the part of the quarter disc in the first quadrant that lies above the line. Its area is the quarter circle area pi by four minus the area of the right triangle with legs one and one, which is one half. The area is therefore pi by four minus one half square units.

Q10 (CBSE Board 2023, 5 marks)

Apply integration to the area of the triangle whose vertices are minus one comma two, one comma five and three comma four.

Answer: The sides are the segments joining the vertices. The lower boundary is the line joining minus one comma two and three comma four, whose equation is y equal to half x plus five halves. The upper boundary is the line joining minus one comma two and one comma five for x from minus one to one, and the line joining one comma five and three comma four for x from one to three. Integrating upper minus lower on the two pieces and adding gives four square units. The determinant formula for the area of the triangle with these vertices also gives four, which checks the result.

A useful habit in the answer is to name the boundary curves and the limits before writing the integral. The examiner sees the setup and rewards it even when the final arithmetic slips, and the setup is also where most of the reasoning marks sit.