Class 12 Mathematics Chapter 7 Integrals – Extra Questions with Answers

This set of questions on Integrals runs through standard antiderivatives and techniques such as integration by parts, the reverse process to differentiation.

Last Updated: September 23, 2026

Very Short Answer Questions (1 mark)

Q1. Evaluate ∫x³dx.
Ans: x⁴/4+C.

Q2. Evaluate ∫sec²x dx.
Ans: tan x+C.

Q3. Evaluate ∫₀¹2x dx.
Ans: [x²]₀¹=1−0=1.

Q4. State the formula for integration by parts.
Ans: ∫u dv=uv−∫v du.

Q5. Evaluate ∫1/x dx.
Ans: log|x|+C.

Short Answer Questions (2–3 marks)

Q6. Evaluate ∫x cos x dx using integration by parts.
Ans: Let u=x, dv=cos x dx, so du=dx, v=sin x. ∫x cos x dx=x sin x−∫sin x dx=x sin x+cos x+C.

Q7. Evaluate ∫₀^π/2sin x dx.
Ans: [−cos x]₀^π/2=−cos(π/2)−(−cos 0)=0−(−1)=1.

Q8. Evaluate ∫2x/(1+x²) dx using substitution.
Ans: Let t=1+x², dt=2x dx. Integral becomes ∫dt/t=log|t|+C=log|1+x²|+C.

Higher-Order Thinking / Application Questions

Q9. Evaluate ∫₀^π/2√(sin x)/[√(sin x)+√(cos x)] dx using the property ∫₀^af(x)dx=∫₀^af(a−x)dx.
Ans: Let I=∫₀^π/2√(sin x)/[√(sin x)+√(cos x)] dx. Applying the property with a=π/2, replace x with (π/2−x): sin(π/2−x)=cos x and cos(π/2−x)=sin x, so I=∫₀^π/2√(cos x)/[√(cos x)+√(sin x)] dx. Call this second expression I again (it equals the same I by the property, just rewritten). Adding the original I and this rewritten I: 2I=∫₀^π/2{√(sin x)/[√(sin x)+√(cos x)]+√(cos x)/[√(cos x)+√(sin x)]}dx=∫₀^π/2[√(sin x)+√(cos x)]/[√(sin x)+√(cos x)] dx=∫₀^π/21 dx=π/2. Therefore 2I=π/2, giving I=π/4. This elegant technique — adding a definite integral to its (a−x)-substituted version to cancel the complicated denominator — is a standard trick for definite integrals with symmetric trigonometric structure.

Q10. Find the area bounded by the curve y=x² and the line y=4 using definite integrals, explaining the setup with a sketch description.
Ans: The parabola y=x² and the horizontal line y=4 intersect where x²=4, i.e., x=−2 and x=2. Between these points, the line y=4 lies above the parabola y=x² (since e.g. at x=0, line value 4>parabola value 0). The bounded region is symmetric about the y-axis, resembling a shape between the upward parabola and the horizontal line capping it. The area is Area=∫₋₂²(4−x²)dx (top curve minus bottom curve, integrated over x). By symmetry, this equals 2∫₀²(4−x²)dx=2[4x−x³/3]₀²=2[(8−8/3)−0]=2[(24−8)/3]=2(16/3)=32/3 square units. This demonstrates the general method for area between curves: identify intersection points as limits, subtract the lower function from the upper function, and integrate (using symmetry to simplify computation where applicable).

Frequently Asked Questions

How would you evaluate the integral of x squared with respect to x?
Using the power rule for integration, the integral is x cubed over 3 plus a constant.

How would you evaluate the definite integral of sin(x) from 0 to pi?
The antiderivative of sin(x) is negative cos(x), so evaluating from 0 to pi gives 1+1 = 2.

Written by Satish

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