What is integration by substitution?

If the integrand contains a function and its derivative, let u = g(x), then du = g'(x)dx. Replace g(x) with u and g'(x)dx with du, integrate, then substitute back.

What is the fundamental theorem of calculus?

If F is an antiderivative of f on [a,b], then ∫[a to b] f(x) dx = F(b) − F(a). The definite integral equals the net area between the curve and the x-axis.

How do you find the area under a curve?

The area bounded by y = f(x), the x-axis, and lines x = a and x = b is A = ∫[a to b] f(x) dx. If the curve goes below the x-axis, take the absolute value.

Integral Calculus Exercise Solutions — Class 11 Business Maths (TN State Board)

Complete worked solutions for the Integral Calculus chapter from the Samacheer Kalvi 11th Business Maths textbook. Covers indefinite integrals, definite integrals, integration by substitution, integration by parts, and area under curves.

How to Use This Page

1. Quick recap first — Read the Key Concepts tables to refresh formulas before attempting problems. 2. Attempt then check — Try each exercise question on your own, then compare with the worked solution here. 3. Focus on the method — Every answer shows each intermediate step. Understanding how we get the answer matters more than the final result. 4. Code-block formulas — Calculations inside text code blocks use plain-text notation (e.g., x^2 for x², sqrt(x) for √x). This makes them easy to copy into notes. 5. Common Mistakes section — Review this before exams to avoid frequent pitfalls.

Key Concepts (Quick Recap)

Standard Integrals

Function f(x)	Integral ∫f(x) dx
xⁿ (n ≠ −1)	xⁿ⁺¹ / (n + 1) + C
1/x	ln\	x\	+ C
eˣ	eˣ + C
aˣ	aˣ / ln(a) + C
sin x	−cos x + C
cos x	sin x + C
sec²x	tan x + C
cosec²x	−cot x + C
sec x tan x	sec x + C
cosec x cot x	−cosec x + C
1 / √(1 − x²)	sin⁻¹x + C
1 / (1 + x²)	tan⁻¹x + C

Integration Rules

Rule	Formula
Constant Multiple	∫k · f(x) dx = k · ∫f(x) dx
Sum Rule	∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx
Difference Rule	∫[f(x) − g(x)] dx = ∫f(x) dx − ∫g(x) dx

Integration by Substitution

If the integrand contains g(x) and g'(x), let u = g(x).
Then du = g'(x) dx.

Replace g(x) with u and g'(x)dx with du.
Integrate with respect to u, then substitute back.

Example: ∫ 2x · (x² + 3)^4 dx
  Let u = x² + 3 → du = 2x dx
  = ∫ u^4 du = u^5 / 5 + C = (x² + 3)^5 / 5 + C

Integration by Parts

∫u dv = u · v − ∫v du

Choose u using LIATE priority:
  L — Logarithmic (ln x, log x)
  I — Inverse trig (sin⁻¹x, tan⁻¹x, ...)
  A — Algebraic (x, x², polynomials)
  T — Trigonometric (sin x, cos x, ...)
  E — Exponential (eˣ, aˣ)

The function earlier in LIATE is chosen as u;
the remaining function is dv.

Definite Integrals

If F'(x) = f(x), then:

  ∫[a to b] f(x) dx = F(b) − F(a)

This gives the net signed area between the curve y = f(x)
and the x-axis from x = a to x = b.

Properties of Definite Integrals

Property	Formula
Reverse limits	∫[a to b] f(x) dx = −∫[b to a] f(x) dx
Zero-width interval	∫[a to a] f(x) dx = 0
Split interval	∫[a to b] f(x) dx = ∫[a to c] f(x) dx + ∫[c to b] f(x) dx
Constant factor	∫[a to b] k·f(x) dx = k · ∫[a to b] f(x) dx
Sum/Difference	∫[a to b] [f(x) ± g(x)] dx = ∫[a to b] f(x) dx ± ∫[a to b] g(x) dx
Even function	∫[−a to a] f(x) dx = 2 · ∫[0 to a] f(x) dx, if f(−x) = f(x)
Odd function	∫[−a to a] f(x) dx = 0, if f(−x) = −f(x)

Area Under Curves

Area between y = f(x), the x-axis, x = a, and x = b:

  A = ∫[a to b] f(x) dx     (when f(x) ≥ 0 on [a, b])

If f(x) dips below the x-axis, use absolute value:

  A = ∫[a to b] |f(x)| dx

Area between two curves y = f(x) and y = g(x):

  A = ∫[a to b] [f(x) − g(x)] dx    (where f(x) ≥ g(x))

I. Choose the Best Answer (MCQs)

1. ∫ (3x² + 2x) dx equals:

(A) x³ + x² + C
(B) 6x + 2 + C
(C) x³ + x²
(D) 3x³ + 2x² + C

Answer(A) ``text ∫ (3x² + 2x) dx = 3 · x³/3 + 2 · x²/2 + C = x³ + x² + C ``

2. ∫₀¹ (2x + 3) dx equals:

(A) 3
(B) 4
(C) 5
(D) 6

Answer(B) ``text ∫₀¹ (2x + 3) dx = [x² + 3x]₀¹ = (1 + 3) − (0 + 0) = 4 ``

3. ∫ e^(3x) dx equals:

(A) e^(3x) + C
(B) (1/3) e^(3x) + C
(C) 3 e^(3x) + C
(D) (1/3) e^(3x)

Answer(B) ``text Let u = 3x → du = 3 dx → dx = du/3 ∫ e^(3x) dx = (1/3) ∫ e^u du = (1/3) e^u + C = (1/3) e^(3x) + C ``

4. ∫ sin²x dx equals:

(A) (x/2) + (sin 2x)/4 + C
(B) (x/2) − (sin 2x)/4 + C
(C) −(x/2) + (sin 2x)/4 + C
(D) (x/2) − (cos 2x)/4 + C

Answer(B) ``

text Using identity: sin²x = (1 − cos 2x) / 2  ∫ sin²x dx = ∫ (1 − cos 2x)/2 dx = (1/2) ∫ 1 dx − (1/2) ∫ cos 2x dx = x/2 − (1/2) · (sin 2x)/2 + C = x/2 − (sin 2x)/4 + C

5. ∫₀²	x − 1	dx equals:

(A) 0
(B) 1
(C) 2
(D) 4

Answer(B) ``

text |x − 1| changes sign at x = 1.  For 0 ≤ x ≤ 1:  x − 1 < 0, so |x − 1| = 1 − x For 1 ≤ x ≤ 2:  x − 1 ≥ 0, so |x − 1| = x − 1  ∫₀² |x − 1| dx = ∫₀¹ (1 − x) dx + ∫₁² (x − 1) dx = [x − x²/2]₀¹ + [x²/2 − x]₁² = (1 − 1/2) + (2 − 1 − 1/2 + 1) = 1/2 + 1/2 = 1

6. If ∫₀ᵃ x dx = 18, then a equals:

(A) 4
(B) 5
(C) 6
(D) 9

Answer(C) ``text ∫₀ᵃ x dx = [x²/2]₀ᵃ = a²/2 = 18 a² = 36 a = 6 ``

7. The value of ∫[−1 to 1] x³ dx is:

(A) 1
(B) −1
(C) 0
(D) 2

Answer(C) ``

text f(x) = x³ is an odd function: f(−x) = (−x)³ = −x³ = −f(x)  ∫[−a to a] f(x) dx = 0 for any odd function.  Therefore ∫[−1 to 1] x³ dx = 0

II. Fill in the Blanks

1. ∫ 5 dx = \_\_\_\_\_\_

5x + C ``text ∫ 5 dx = 5x + C (integral of a constant k is kx + C) ``

2. ∫ (1/x) dx = \_\_\_\_\_\_

ln|x| + C

3. d/dx [∫f(x) dx] = \_\_\_\_\_\_

f(x) ``text Differentiation is the inverse of integration. d/dx [∫f(x) dx] = f(x) ``

4. ∫₀³ x² dx = \_\_\_\_\_\_

9 ``text ∫₀³ x² dx = [x³/3]₀³ = 27/3 − 0 = 9 ``

5. ∫ sec²x dx = \_\_\_\_\_\_

tan x + C

6. If ∫[0 to 2] f(x) dx = 5, then ∫[0 to 2] 3f(x) dx = \_\_\_\_\_\_

15 ``text ∫[0 to 2] 3f(x) dx = 3 · ∫[0 to 2] f(x) dx = 3 · 5 = 15 ``

7. ∫ eˣ dx = \_\_\_\_\_\_

eˣ + C

8. ∫[a to b] f(x) dx + ∫[b to c] f(x) dx = \_\_\_\_\_\_

∫[a to c] f(x) dx ``text This is the interval-splitting property of definite integrals. ``

9. ∫ sin x dx = \_\_\_\_\_\_

−cos x + C

10. In integration by parts, ∫u dv = \_\_\_\_\_\_

uv − ∫v du

11. ∫₀¹ eˣ dx = \_\_\_\_\_\_

e − 1 ``text ∫₀¹ eˣ dx = [eˣ]₀¹ = e¹ − e⁰ = e − 1 ``

12. If f(x) is an odd function, then ∫[−a to a] f(x) dx = \_\_\_\_\_\_

III. True or False

1. ∫ f(x) dx is called the indefinite integral of f(x).

True. The indefinite integral represents the family of all antiderivatives of f(x) and includes an arbitrary constant C.

2. The constant of integration C is added only in definite integrals.

False. The constant of integration C appears in indefinite integrals, not definite integrals. In a definite integral ∫[a to b] f(x) dx, the constant cancels out: F(b) − F(a), so C is not needed.

3. ∫[a to b] f(x) dx = ∫[b to a] f(x) dx.

False. Reversing the limits changes the sign: ∫[a to b] f(x) dx = −∫[b to a] f(x) dx.

4. Integration is the reverse process of differentiation.

True. If d/dx [F(x)] = f(x), then ∫f(x) dx = F(x) + C. Integration undoes differentiation (up to a constant).

5. The area under the curve y = x² from x = 0 to x = 2 is given by ∫₀² x² dx.

True. Since x² ≥ 0 for all real x, the area between y = x², the x-axis, x = 0 and x = 2 equals ∫₀² x² dx = 8/3 square units.

IV. Match the Following

Match the integral in Column A with the result in Column B.

#	Column A (Integral)	Column B (Result)
1	∫ cos x dx	(a) x²/2 + C
2	∫ x dx	(b) ln\	x\	+ C
3	∫ (1/x) dx	(c) −cot x + C
4	∫ cosec²x dx	(d) sin x + C
5	∫ 0 dx	(e) C (constant)
6	∫ sec x tan x dx	(f) sec x + C

Answer

#	Match
1	∫ cos x dx → (d) sin x + C
2	∫ x dx → (a) x²/2 + C
3	∫ (1/x) dx → **(b) ln\	x\	+ C**
4	∫ cosec²x dx → (c) −cot x + C
5	∫ 0 dx → (e) C
6	∫ sec x tan x dx → (f) sec x + C

V. Answer Briefly (2-Mark Questions)

1. State the power rule of integration with an example.

The power rule states that for any n ≠ −1: ``text ∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C `` Example: ∫ x⁴ dx = x⁵ / 5 + C

2. Evaluate: ∫ (2x + 3) dx

``text ∫ (2x + 3) dx = 2 · ∫ x dx + ∫ 3 dx = 2 · x²/2 + 3x + C = x² + 3x + C ``

3. What is the geometric meaning of a definite integral?

The definite integral ∫[a to b] f(x) dx represents the net signed area between the curve y = f(x) and the x-axis from x = a to x = b. Areas above the x-axis are positive and areas below are negative.

4. Evaluate: ∫₀¹ (2x + 1) dx

``text ∫₀¹ (2x + 1) dx = [x² + x]₀¹ = (1 + 1) − (0 + 0) = 2 ``

VI. Answer in Detail (5-Mark Questions)

1. Evaluating Indefinite Integrals Using Substitution

Evaluate: ∫ x · √(x² + 4) dx

Step 1: Identify the substitution.
  The integrand has √(x² + 4) and its derivative 2x (up to a constant).
  Let  u = x² + 4

Step 2: Compute du.
  du = 2x dx
  x dx = du / 2

Step 3: Substitute into the integral.
  ∫ x · √(x² + 4) dx = ∫ √(u) · (du / 2)
                      = (1/2) ∫ u^(1/2) du

Step 4: Integrate with respect to u.
  = (1/2) · u^(3/2) / (3/2) + C
  = (1/2) · (2/3) · u^(3/2) + C
  = (1/3) · u^(3/2) + C

Step 5: Substitute back u = x² + 4.
  = (1/3) · (x² + 4)^(3/2) + C

Verification (by differentiation):

  d/dx [(1/3)(x² + 4)^(3/2)]
    = (1/3) · (3/2) · (x² + 4)^(1/2) · 2x
    = x · (x² + 4)^(1/2)
    = x · √(x² + 4)  ✓

Another example: ∫ (3x²) / (x³ + 1) dx

Step 1: Let u = x³ + 1  (denominator)
Step 2: du = 3x² dx  →  3x² dx = du
Step 3: ∫ (3x²) / (x³ + 1) dx = ∫ (1/u) du
Step 4: = ln|u| + C
Step 5: = ln|x³ + 1| + C

2. Evaluating a Definite Integral Step-by-Step

Evaluate: ∫₁³ (2x² + 3x − 1) dx

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

  Let F(x) = 2x³/3 + 3x²/2 − x

Step 2: Apply the limits.
  F(3) = 2(27)/3 + 3(9)/2 − 3
       = 54/3 + 27/2 − 3
       = 18 + 13.5 − 3
       = 28.5

  F(1) = 2(1)/3 + 3(1)/2 − 1
       = 2/3 + 3/2 − 1

  Common denominator = 6:
       = 4/6 + 9/6 − 6/6
       = 7/6

Step 3: Subtract.
  ∫₁³ (2x² + 3x − 1) dx = F(3) − F(1)
                          = 28.5 − 7/6
                          = 57/2 − 7/6

  Common denominator = 6:
                          = 171/6 − 7/6
                          = 164/6
                          = 82/3
                          ≈ 27.33

Another example: ∫₀^(π/2) (sin x + cos x) dx

Step 1: Find the antiderivative.
  ∫ (sin x + cos x) dx = −cos x + sin x + C

  Let F(x) = −cos x + sin x

Step 2: Evaluate at upper limit.
  F(π/2) = −cos(π/2) + sin(π/2)
         = −0 + 1
         = 1

Step 3: Evaluate at lower limit.
  F(0) = −cos 0 + sin 0
       = −1 + 0
       = −1

Step 4: Subtract.
  ∫₀^(π/2) (sin x + cos x) dx = F(π/2) − F(0)
                                = 1 − (−1)
                                = 2

3. Finding Area Under a Curve

Find the area bounded by y = x², the x-axis, and the lines x = 0 and x = 3.

Step 1: Identify the region.
  The curve is y = x², which is always ≥ 0.
  The boundaries are x = 0 (y-axis) and x = 3.

Step 2: Set up the integral.
  Area = ∫₀³ x² dx

Step 3: Compute the antiderivative.
  ∫ x² dx = x³/3

Step 4: Apply limits.
  Area = [x³/3]₀³
       = 27/3 − 0/3
       = 9 square units

Conclusion: The area under y = x² from x = 0 to x = 3 is 9 sq. units.

Find the area enclosed between y = x and y = x².

Step 1: Find the points of intersection.
  Set x = x²
  x² − x = 0
  x(x − 1) = 0
  x = 0  or  x = 1

  The curves intersect at x = 0 and x = 1.

Step 2: Determine which curve is on top.
  For 0 < x < 1:  x > x²  (since 0 < x < 1)
  So y = x is the upper curve and y = x² is the lower curve.

Step 3: Set up the area integral.
  Area = ∫₀¹ (x − x²) dx

Step 4: Compute.
  = [x²/2 − x³/3]₀¹
  = (1/2 − 1/3) − (0 − 0)
  = (3/6 − 2/6)
  = 1/6

Conclusion: The area between y = x and y = x² is 1/6 square units.

Common Mistakes to Avoid

1. Forgetting the constant of integration — Every indefinite integral must include "+ C". Without it, you have only one antiderivative, not the entire family. Definite integrals do not need C (it cancels).

2. Wrong power rule for n = −1 — The power rule ∫xⁿ dx = xⁿ⁺¹/(n+1) fails when n = −1 (division by zero). Use ∫(1/x) dx = ln	x	+ C instead.

3. Swapping definite integral limits — ∫[a to b] f(x) dx = −∫[b to a] f(x) dx. If you reverse the limits, the sign flips. Always check the order.

4. Incorrect substitution choices — Choose u so that du appears in the integrand (or a constant multiple of it). If the derivative of u is nowhere in the integrand, substitution will not simplify.

5. Wrong u/dv choice in integration by parts — Use LIATE priority. Choosing dv as the algebraic part and u as the exponential/trig part makes the problem harder, not easier.

6. Missing absolute value in ln — ∫(1/x) dx = ln	x	+ C, not ln(x) + C. The absolute value ensures the domain covers negative x values.

7. Not splitting the integral at sign changes — When finding area under a curve that crosses the x-axis, the definite integral gives signed area. Split the integral at each x-intercept and take absolute values for actual geometric area.

8. Confusing differentiation with integration rules — ∫ sin x dx = −cos x + C, not cos x + C. The derivative of cos x is −sin x, so the integral of sin x is −cos x. Memorize standard integrals separately from derivative rules.

Exam Checklist

Before walking into the exam, verify each item:

[ ] I can state and apply the power rule of integration: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C
[ ] I know the special case n = −1: ∫(1/x) dx = ln|x| + C
[ ] I can recall all 12 standard integrals from the formula table (exponential, trig, inverse trig)
[ ] I can apply the constant multiple and sum/difference rules fluently
[ ] I can identify when substitution is needed and choose the correct u
[ ] I can apply integration by parts using the LIATE rule for choosing u
[ ] I can evaluate definite integrals by finding antiderivatives and applying limits
[ ] I know the properties of definite integrals (reverse limits, split interval, even/odd functions)
[ ] I can compute the area under a curve and area between two curves
[ ] I remember to add + C for indefinite integrals and omit C for definite integrals

Frequently Asked Questions

∫xⁿ dx = xⁿ⁺¹/(n+1) + C, for n ≠ −1. When n = −1, ∫x⁻¹ dx = ∫(1/x) dx = ln|x| + C.

Integral Calculus Exercise Solutions | KalviGavani