Question 1

If ∫₀ᵃ f(x) dx = a² - 2a + 3, then f(a) is equal to:

Accepted Answer

We use the Leibniz rule (differentiation under the integral sign), which is a consequence of the fundamental theorem of calculus.

If F(a) = ∫₀ᵃ f(x) dx, then dF/da = f(a).

Given: F(a) = a² - 2a + 3

Differentiate both sides with respect to a:
f(a) = dF/da = d/da(a² - 2a + 3)
     = 2a - 2

This follows directly from the fundamental theorem: the derivative of a definite integral with respect to its upper limit equals the integrand evaluated at that limit.

Question 2

Evaluate the definite integral ∫₀^(π/2) sin(x) dx.

Accepted Answer

We use the fundamental theorem of calculus. The antiderivative of sin(x) is -cos(x). By the fundamental theorem: ∫₀^(π/2) sin(x) dx = [-cos(x)]₀^(π/2) = -cos(π/2) - (-cos(0)) = -0 - (-1) = 0 + 1 = 1.

Question 3

The area enclosed between the curve y = x and the curve y = x² from x = 0 to x = 1 is:

Accepted Answer

First, determine which curve is above the other on [0, 1]. At x = 0.5: y = x gives 0.5, and y = x² gives 0.25. So y = x is above y = x².

Area = ∫₀¹ (x - x²) dx
     = [x²/2 - x³/3]₀¹
     = (1/2 - 1/3) - (0 - 0)
     = 3/6 - 2/6
     = 1/6

We use the formula: Area between curves = ∫ₐᵇ (upper curve - lower curve) dx.

Question 4

Find the area enclosed between the curve y = x² and the line y = 4 from x = -2 to x = 2.

Accepted Answer

The line y = 4 is above the parabola y = x² in the interval [-2, 2] (since x² ≤ 4 for |x| ≤ 2). Area = ∫₋₂² [4 - x²] dx. Step 1: Find antiderivative: ∫[4 - x²] dx = 4x - x³/3. Step 2: Apply FTC: [4x - x³/3]₋₂² = [4(2) - 2³/3] - [4(-2) - (-2)³/3] = [8 - 8/3] - [-8 - (-8/3)] = [8 - 8/3] - [-8 + 8/3] = [24/3 - 8/3] - [-24/3 + 8/3] = 16/3 - (-16/3) = 16/3 + 16/3 = 32/3.

Question 5

Evaluate ∫₁² (3x² + 2x) dx.

Accepted Answer

We find the antiderivative and apply the fundamental theorem. Step 1: Antiderivative of 3x² + 2x is x³ + x². Step 2: Apply FTC: [x³ + x²]₁² = (2³ + 2²) - (1³ + 1²) = (8 + 4) - (1 + 1) = 12 - 2 = 10.

Question 6

The rate of change of the volume of a sphere with respect to its radius is dV/dr. If V = (4/3)πr³, find dV/dr at r = 2.

Accepted Answer

We differentiate V with respect to r using the power rule. Step 1: V = (4/3)πr³. Step 2: dV/dr = (4/3)π · 3r² = 4πr². Step 3: At r = 2: dV/dr = 4π(2)² = 4π · 4 = 16π.

Question 7

Evaluate ∫₀¹ e^x dx.

Accepted Answer

We use the fact that the antiderivative of e^x is e^x itself. Step 1: ∫e^x dx = e^x + C. Step 2: Apply FTC: [e^x]₀¹ = e¹ - e⁰ = e - 1.

Question 8

Find the area enclosed between the curve y = x² and the line y = 2x in the first quadrant.

Accepted Answer

Step 1: Find intersection points by solving x² = 2x.
x² - 2x = 0
x(x - 2) = 0
x = 0 or x = 2

At x = 0: y = 0; at x = 2: y = 4. Intersection points are (0, 0) and (2, 4).

Step 2: Determine which curve is above. At x = 1: parabola gives y = 1, line gives y = 2. So the line y = 2x is above y = x².

Step 3: Set up the definite integral:
Area = ∫₀² (2x - x²) dx

Step 4: Integrate:
= [x² - x³/3]₀²
= (4 - 8/3) - (0 - 0)
= 12/3 - 8/3
= 4/3

The area is 4/3 square units.

Question 9

Evaluate ∫₀¹ (e^x) dx.

Accepted Answer

Step 1: Find the antiderivative of e^x.
The antiderivative of e^x is e^x (since d/dx(e^x) = e^x).

Step 2: Apply the fundamental theorem of calculus.
∫₀¹ e^x dx = [e^x]₀¹
= e¹ - e⁰
= e - 1

The answer is e - 1, which is approximately 1.718.

Question 10

Find the area under the curve y = x² between x = 0 and x = 2.

Accepted Answer

The area under a curve is given by the definite integral ∫ₐᵇ f(x) dx.

Area = ∫₀² x² dx
     = [x³/3]₀²
     = (2³/3) - (0³/3)
     = 8/3 - 0
     = 8/3

We apply the power rule for integration: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, then evaluate at the boundaries using the fundamental theorem.

Question 11

The area of the region bounded by the curve y = e^x, the x-axis, and the lines x = 0 and x = 1 is:

Accepted Answer

We need to find the area between the curve y = e^x, the x-axis (y = 0), and the vertical lines x = 0 and x = 1.

Since e^x > 0 for all x, the curve lies entirely above the x-axis on [0, 1].

Area = ∫₀¹ e^x dx

The antiderivative of e^x is e^x.

Area = [e^x]₀¹

= e¹ - e⁰

= e - 1

Question 12

The area enclosed between the curve y = x² and the line y = 2x is:

Accepted Answer

Step 1: Find intersection points of y = x² and y = 2x.
  x² = 2x
  x² - 2x = 0
  x(x - 2) = 0
  x = 0 or x = 2

Step 2: Determine which function is above the other in [0, 2].
  At x = 1: y = x² gives 1, y = 2x gives 2. So 2x > x² on [0, 2].

Step 3: Set up the area integral.
  Area = ∫₀² (2x - x²) dx

Step 4: Integrate.
  ∫₀² (2x - x²) dx = [x² - x³/3]₀²

Step 5: Evaluate at limits.
  At x = 2: 2² - 2³/3 = 4 - 8/3 = 12/3 - 8/3 = 4/3
  At x = 0: 0 - 0 = 0

Step 6: Compute the area.
  Area = 4/3 - 0 = 4/3 square units

Question 13

If ∫₀^a (3x² + 2x) dx = 10, then the value of a is:

Accepted Answer

Step 1: Find the antiderivative of 3x² + 2x.
  ∫(3x² + 2x) dx = x³ + x² + C

Step 2: Apply the Fundamental Theorem of Calculus.
  ∫₀^a (3x² + 2x) dx = [x³ + x²]₀^a

Step 3: Evaluate at the limits.
  = (a³ + a²) - (0³ + 0²)
  = a³ + a²

Step 4: Set equal to 10 and solve for a.
  a³ + a² = 10
  a²(a + 1) = 10

Step 5: Test integer and simple rational values.
  If a = 2: 2²(2 + 1) = 4(3) = 12 ✗
  If a = 1.5: (1.5)²(2.5) = 2.25(2.5) = 5.625 ✗
  If a = 1.8: (1.8)²(2.8) = 3.24(2.8) = 9.072 ✗
  If a = 1.9: (1.9)²(2.9) = 3.61(2.9) = 10.469 ✗
  If a = 1.85: (1.85)²(2.85) ≈ 3.4225(2.85) ≈ 9.754 ✗
  If a = 1.87: (1.87)²(2.87) ≈ 3.4969(2.87) ≈ 10.036 ✓

Alternatively, solving a³ + a² - 10 = 0 numerically or by inspection:
  a ≈ 1.8696 (real root)

For NDA purposes, if exact answer is required, we verify: a must satisfy a³ + a² = 10.

Note: The problem expects a = 2 as the closest integer, but exact solution is a ≈ 1.87 or the real root of a³ + a² - 10 = 0.

Question 14

Evaluate: ∫₁² (1/x + e^x) dx

Accepted Answer

Step 1: Identify the antiderivatives of each term.
  ∫(1/x) dx = ln|x| + C
  ∫e^x dx = e^x + C

Step 2: Combine the antiderivatives.
  ∫(1/x + e^x) dx = ln|x| + e^x + C

Step 3: Apply the Fundamental Theorem of Calculus with limits 1 to 2.
  ∫₁² (1/x + e^x) dx = [ln|x| + e^x]₁²

Step 4: Evaluate at the upper limit x = 2.
  ln(2) + e²

Step 5: Evaluate at the lower limit x = 1.
  ln(1) + e¹ = 0 + e = e

Step 6: Subtract lower limit from upper limit.
  ∫₁² (1/x + e^x) dx = (ln(2) + e²) - (0 + e)
  = ln(2) + e² - e
  = ln(2) + e(e - 1)

Question 15

The value of ∫₀^(π/2) sin³(x) dx is:

Accepted Answer

Step 1: Use the reduction formula or trigonometric identity for sin³(x).
  sin³(x) = sin(x)·sin²(x) = sin(x)·(1 - cos²(x))

Step 2: Expand the integrand.
  ∫sin³(x) dx = ∫sin(x)·(1 - cos²(x)) dx
  = ∫sin(x) dx - ∫sin(x)·cos²(x) dx

Step 3: Evaluate the first integral.
  ∫sin(x) dx = -cos(x) + C

Step 4: Evaluate the second integral using substitution u = cos(x), du = -sin(x) dx.
  ∫sin(x)·cos²(x) dx = -∫cos²(x)·(-sin(x)) dx = -∫u² du = -u³/3 = -cos³(x)/3 + C

Step 5: Combine the antiderivatives.
  ∫sin³(x) dx = -cos(x) + cos³(x)/3 + C

Step 6: Apply limits from 0 to π/2.
  ∫₀^(π/2) sin³(x) dx = [-cos(x) + cos³(x)/3]₀^(π/2)

Step 7: Evaluate at x = π/2.
  -cos(π/2) + cos³(π/2)/3 = -0 + 0/3 = 0

Step 8: Evaluate at x = 0.
  -cos(0) + cos³(0)/3 = -1 + 1/3 = -2/3

Step 9: Subtract lower limit from upper limit.
  ∫₀^(π/2) sin³(x) dx = 0 - (-2/3) = 2/3

Question 16

Evaluate the definite integral ∫₀^(π/2) x·sin(x) dx.

Accepted Answer

We use integration by parts with u = x and dv = sin(x)dx.

Then du = dx and v = -cos(x).

By integration by parts: ∫ u dv = uv - ∫ v du

∫₀^(π/2) x·sin(x) dx = [x·(-cos(x))]₀^(π/2) - ∫₀^(π/2) (-cos(x)) dx

= [-x·cos(x)]₀^(π/2) + ∫₀^(π/2) cos(x) dx

= [-x·cos(x) + sin(x)]₀^(π/2)

At x = π/2: -(π/2)·cos(π/2) + sin(π/2) = -(π/2)·0 + 1 = 1

At x = 0: -0·cos(0) + sin(0) = 0

Therefore: ∫₀^(π/2) x·sin(x) dx = 1 - 0 = 1

Question 17

The value of ∫₀^(π/4) tan(x) dx is:

Accepted Answer

We need to evaluate ∫₀^(π/4) tan(x) dx.

Recall that tan(x) = sin(x)/cos(x).

∫ tan(x) dx = ∫ sin(x)/cos(x) dx

Use substitution: let u = cos(x), then du = -sin(x) dx.

∫ sin(x)/cos(x) dx = -∫ du/u = -ln|u| + C = -ln|cos(x)| + C

Alternatively, this can be written as ln|sec(x)| + C.

Now evaluate the definite integral:

∫₀^(π/4) tan(x) dx = [-ln|cos(x)|]₀^(π/4)

= -ln|cos(π/4)| - (-ln|cos(0)|)

= -ln(1/√2) + ln(1)

= -ln(1/√2) + 0

= -ln(2^(-1/2))

= -(-1/2)ln(2)

= (1/2)ln(2)

= ln(√2)

Alternatively: -ln(cos(π/4)) + ln(cos(0)) = -ln(√2/2) + ln(1) = -ln(√2/2) = ln(2/√2) = ln(√2) = (1/2)ln(2).

Question 18

Find the area enclosed between the curve y = x² and the line y = 2x in the first quadrant.

Accepted Answer

First, find intersection points by solving x² = 2x:
x² - 2x = 0
x(x - 2) = 0
x = 0 or x = 2

At x = 0: y = 0. At x = 2: y = 4.
Intersection points: (0, 0) and (2, 4).

For 0 ≤ x ≤ 2, the line y = 2x lies above the parabola y = x².

Area = ∫₀² (2x - x²) dx

= [x² - x³/3]₀²

= (4 - 8/3) - (0 - 0)

= 12/3 - 8/3

= 4/3

Question 19

Evaluate ∫₁² (3x² + 2/x) dx.

Accepted Answer

Split the integral:
∫₁² (3x² + 2/x) dx = ∫₁² 3x² dx + ∫₁² (2/x) dx

For the first integral:
∫₁² 3x² dx = [x³]₁² = 2³ - 1³ = 8 - 1 = 7

For the second integral:
∫₁² (2/x) dx = [2·ln(x)]₁² = 2·ln(2) - 2·ln(1) = 2·ln(2) - 0 = 2·ln(2)

Therefore:
∫₁² (3x² + 2/x) dx = 7 + 2·ln(2)

Question 20

The area under the curve y = e^x from x = 0 to x = ln(2) is:

Accepted Answer

We need to evaluate ∫₀^(ln 2) e^x dx.

The antiderivative of e^x is e^x.

∫₀^(ln 2) e^x dx = [e^x]₀^(ln 2)

= e^(ln 2) - e^0

= 2 - 1

= 1

NDA Definite Integrals & Applications

Definite Integrals & Applications— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Definite Integrals & Applications — Practice Questions

60-Second Revision — Definite Integrals & Applications