ilovepdf_merged (6).pdf). Content covers sections 31.1–31.x.The definite integral ∫_a^b f(x) dx equals F(b)−F(a) when F′=f (Fundamental Theorem), and represents net signed area under y=f(x) from a to b.
Area between curve and x-axis may need absolute value if f changes sign. Area between two curves: ∫ |f−g| over the interval of intersection projections.
Most exam-important points from this chapter:
Start every answer with the key definition or standard form from Definite Integrals.
Copy the formula, then substitute values; never jump to the number alone.
Check domain, quadrant, non-zero denominators, and applicability of the theorem.
Memorise and apply: = 0 if f is an odd function of x.
Memorise and apply: 31.1 DEFINITE INTEGRAL AS A LIMIT OF SUM
English questions from NIOS Mathematics (311) public / sample papers (Sample 2024, Apr 2024, Oct 2024, Apr 2025; board-style where scanned papers had no extractable text). Mapped exclusively to this chapter — no cross-chapter duplicates. Use Model Answer for the marking key and Explanation for working.
4 question(s) · Sources: Board-style, Board-style FTC, Board-style property, Board-style reverse limits
PYQ1. Evaluate ∫_0^2 3x² dx.
Model Answer
1) 8
Explanation
1) [x³]_0^2 = 8.
PYQ2. Evaluate ∫_0^π sin x dx.
Model Answer
1) 2
Explanation
1) [−cos x]_0^π = −(−1) − (−1) = 1+1=2.
PYQ3. Show that I = ∫_0^{π/2} [sin x /(sin x + cos x)] dx equals π/4.
Model Answer
1) I = π/4
Explanation
1) Using x → π/2−x gives I=∫ cos/(sin+cos). Adding: 2I=∫_0^{π/2}1 dx=π/2
2) I=π/4.
PYQ4. If ∫_1^4 f(x) dx = 7, find ∫_4^1 f(x) dx.
Model Answer
1) −7
Explanation
1) ∫_a^b = −∫_b^a.
10 English exam-style problems for this chapter (NIOS Mathematics 311). Each question states the formula, gives full step-by-step working with the final answer, plus formal/easy explanations. English only.
Evaluate ∫_0^2 (3x²) dx.
1) F=x³; F(2)−F(0)=8−0=8.
Answer: 8
Fundamental theorem with antiderivative x³.
Cube of 2 is 8.
Lower limit 0 vanishes.
3x² → x³ not 3x³.
If ∫_1^4 f = 7, find ∫_4^1 f.
1) ∫_4^1 f = −7.
Answer: −7
Reversing limits multiplies by −1.
Swap ends, put a minus.
Independent of the path of f as long as integrable.
Do not keep +7.
Compute ∫_0^π sin x dx.
1) [−cos x]_0^π = −cosπ − (−cos0) = −(−1) + 1 = 2.
Answer: 2
Net area under one arch of sine from 0 to π is 2.
−cos at π is +1; minus −1 at 0 gives 2.
Total geometric area same here since sin≥0 on [0,π].
Careful with −(−cos0).
Using the property, show I=∫_0^{π/2} sin x /(sin x + cos x) dx equals π/4.
1) I=∫_0^{π/2} sin x/(sin x+cos x) dx.
2) Also I=∫_0^{π/2} cos x/(sin x+cos x) dx by x→π/2−x.
3) Add: 2I=∫_0^{π/2} 1 dx=π/2 ⇒ I=π/4.
Answer: π/4
The substitution x→a+b−x symmetrises sine and cosine.
Two expressions for I add to the integral of 1.
Classic NIOS/board trick for this integrand.
Limits 0 to π/2 are essential for the swap.
Evaluate ∫_1^2 (2x + 3) dx.
1) [x² + 3x]_1^2 = (4+6) − (1+3) = 10−4=6.
Answer: 6
Antiderivative x²+3x evaluated between limits.
At 2:10; at 1:4; difference 6.
Can split as 2∫x + 3∫1.
Arithmetic at the bounds.
Find the total area between y=x and the x-axis from x=−1 to x=2.
1) ∫_−1^0 (−x)dx + ∫_0^2 x dx = [−x²/2]_−1^0 + [x²/2]_0^2 = (0−(−1/2)) + (2−0) = 0.5+2=2.5.
Answer: 5/2 (or 2.5)
Total area uses |x|, splitting at the root x=0.
Triangle areas 1/2 + 2 = 2.5.
Net ∫_−1^2 x dx = 1.5 would be wrong for total area.
Split where the function changes sign.
Evaluate ∫_1^4 2x dx.
1) F=x²; F(4)−F(1)=16−1=15.
Answer: 15
Fundamental theorem of calculus.
Antiderivative x² from 1 to 4.
No +C needed.
Check each algebraic step carefully.
Find the area under y = x from x = 0 to x = 3.
1) ∫_0^3 x dx = [x²/2]_0^3 = 9/2.
Answer: 9/2
Definite integral of a non-negative function is area.
Triangle area ½·3·3=9/2.
Geometry checks the integral.
Check each algebraic step carefully.
If ∫_0^2 f(x) dx = 7, find ∫_2^0 f(x) dx.
1) Reversing limits changes sign: −7.
Answer: −7
Orientation of the interval.
Swap bounds ⇒ minus.
Same integrand.
Check each algebraic step carefully.
Evaluate ∫_−1^1 x² dx.
1) x² even: 2∫_0^1 x² dx = 2[x³/3]_0^1 = 2/3.
Answer: 2/3
Even function property on symmetric interval.
Or direct: [x³/3]_−1^1 = 1/3 − (−1/3)=2/3.
Odd integrands over symmetric limits give 0.
Check each algebraic step carefully.