Class 12 Mathematics Chapter 7 Integrals – Extra Questions with Answers

Extra practice questions for Class 12 Maths Chapter 7 (Integrals), beyond the textbook. These Class 12 Mathematics Chapter 7 important questions are handy for last-minute exam practice.

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).

Written by Satish

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