JEE Main Mathematics · Chapter 10

Integral Calculus

Standard integrals, substitution, integration by parts, partial fractions, definite integrals and area under curves.

NCERT Class 12JEE Main: 2–3 Qs per shift 7 min read

1. Chapter at a Glance

Integration is the reverse of differentiation. An indefinite integral finds a function whose derivative is given; a definite integral gives a number that represents the signed area under a curve. Integral calculus is the largest single block of JEE Main mathematics — expect 2 or 3 questions every shift, especially on properties of definite integrals and area between curves.

What JEE asks most
  • Definite integrals solved by properties (King's property, odd/even, periodicity).
  • Integrals of the form ∫eˣ(f + f′)dx, partial fractions and trigonometric substitutions.
  • Integrals with [x], |x| and other piecewise functions.
  • Area bounded by curves such as parabolas and lines; limits as Riemann sums.

2. Standard Indefinite Integrals

If F′(x) = f(x), then ∫f(x) dx = F(x) + C, where C is the constant of integration.

IntegralResult
∫xⁿ dx (n ≠ −1)xⁿ⁺¹/(n + 1)
∫1/x dxln|x|
∫eˣ dx, ∫aˣ dxeˣ, aˣ/ln a
∫sin x dx, ∫cos x dx−cos x, sin x
∫tan x dxln|sec x|
∫sec x dxln|sec x + tan x|
∫sec²x dxtan x
∫dx/(x² + a²)(1/a) tan−1(x/a)
∫dx/√(a² − x²)sin−1(x/a)
∫dx/(x² − a²)(1/2a) ln|(x − a)/(x + a)|
∫dx/√(x² ± a²)ln|x + √(x² ± a²)|
∫√(a² − x²) dx(x/2)√(a² − x²) + (a²/2) sin−1(x/a)

(Add + C to every indefinite integral.)

3. Methods of Integration

Substitution

If the integrand contains a function and its derivative, substitute t = that function. Example: ∫2x cos(x²) dx, put t = x², dt = 2x dx ⇒ sin(x²) + C. In particular ∫f′(x)/f(x) dx = ln|f(x)| + C.

Integration by parts

∫u v dx = u∫v dx − ∫(u′ ∫v dx) dx

Choose u by the ILATE order: Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential — the function that comes first is u.

Special result
∫eˣ [f(x) + f′(x)] dx = eˣ f(x) + C

Spot this pattern: ∫eˣ(sin x + cos x) dx = eˣ sin x + C; ∫eˣ(1/x − 1/x²) dx = eˣ/x + C.

Partial fractions

For a proper rational function, split the denominator's factors: A/(x − a) for linear factors, B/(x − a)² … for repeated factors, (Cx + D)/(x² + px + q) for irreducible quadratics.

Trigonometric integrals

  • Use identities to reduce powers: sin²x = (1 − cos 2x)/2, cos²x = (1 + cos 2x)/2.
  • ∫dx/(a + b cos x) or ∫dx/(a + b sin x): put t = tan(x/2).
  • ∫(a sin x + b cos x)/(c sin x + d cos x) dx: write numerator = A(denominator) + B(derivative of denominator).

4. Definite Integrals and their Properties

∫ab f(x) dx = F(b) − F(a)
PropertyStatement
Reversal∫ab f = −∫ba f
Splitting∫ab f = ∫ac f + ∫cb f
King's property∫ab f(x) dx = ∫ab f(a + b − x) dx
Queen's property∫02a f(x) dx = ∫0a [f(x) + f(2a − x)] dx
Even function∫−aa f = 2∫0a f
Odd function∫−aa f = 0
Periodic (period T)∫0nT f = n∫0T f
Leibniz ruled/dx ∫g(x)h(x) f(t)dt = f(h)h′ − f(g)g′

Definite integral as a limit of a sum

limn→∞ (1/n) Σr=1n f(r/n) = ∫01 f(x) dx

Replace r/n by x, 1/n by dx and the limits of r/n by the limits of x.

Integrals with [x] and |x|

Split the range at the points where [x] jumps (integers) or where the expression inside |·| changes sign.

5. Area Under and Between Curves

Area between y = f(x), the x-axis, and x = a, x = b is ∫ab |f(x)| dx. Area between two curves (upper minus lower):

Area = ∫ab [f(x) − g(x)] dx, where f ≥ g on [a, b]
y = xy = x² (1, 1) Shaded area = 1/6 x = 1
Between x = 0 and x = 1 the line y = x lies above y = x², so the shaded area is ∫(x − x²)dx = 1/2 − 1/3 = 1/6.
  • Area inside the parabola y² = 4ax cut off by its latus rectum: 8a²/3.
  • Area of an ellipse x²/a² + y²/b² = 1: πab.
  • Area between y² = 4ax and x² = 4by: 16ab/3.

6. Formula Master Table

ResultFormula
By parts∫uv = u∫v − ∫(u′∫v)
eˣ pattern∫eˣ(f + f′) = eˣf
Log pattern∫f′/f = ln|f|
King's propertyf(x) → f(a + b − x)
∫0π/2 sinⁿx/(sinⁿx + cosⁿx)π/4
∫0π/2 ln sin x dx−(π/2) ln 2
Wallis: ∫0π/2 sin²x dxπ/4
Riemann sumlim (1/n)Σf(r/n) = ∫01f
Area of ellipseπab
Parabola–latus rectum area8a²/3

7. Mind Map

Integral Calculus
Indefinite
  • Standard results
  • Substitution
  • By parts (ILATE)
  • Partial fractions
Definite
  • F(b) − F(a)
  • King, Queen
  • Odd/even, periodic
  • Leibniz rule
Limit of a sum
  • r/n → x, 1/n → dx
Area
  • Upper − lower
  • Split at intersections
  • Parabola, ellipse results

8. Tips and Tricks

Shortcuts
  • Whenever a definite integral looks hard, try King's property first: add the original and the transformed integrals — often the sum simplifies to a constant.
  • Check for odd functions on symmetric limits: x³cos x, sin³x, x·e−x² all integrate to 0 over [−a, a].
  • For area problems, always draw a quick sketch and find the intersection points first.
  • Differentiate the options when you are stuck with an indefinite integral MCQ — the correct option will differentiate back to the integrand.
Mistakes students make
  • Forgetting to change the limits when substituting in a definite integral.
  • Taking area as ∫f dx when part of the curve lies below the x-axis — negative parts must be taken with modulus.
  • Writing ∫1/x dx = ln x instead of ln|x|.

9. Important One-liners

  • The derivative of ∫ax f(t) dt with respect to x is f(x).
  • ∫0a f(x) dx = ∫0a f(a − x) dx.
  • If f is odd, ∫−aa f = 0; if even, it is twice ∫0a.
  • ∫0π x f(sin x) dx = (π/2)∫0π f(sin x) dx.
  • ∫0n [x] dx = n(n − 1)/2 for a positive integer n.
  • The area enclosed by |x| + |y| = a is 2a².
  • Integration by parts twice is needed for ∫eˣ sin x dx; the integral reappears and can be solved for.

10. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

∫ dx/(x² + 4) equals:

  1. tan−1(x/2) + C
  2. (1/2) tan−1(x/2) + C
  3. 2 tan−1(x/2) + C
  4. (1/4) tan−1(x/4) + C
Show solution

Using ∫dx/(x² + a²) = (1/a) tan−1(x/a) with a = 2.

Answer: (B)

Q2Medium

∫0π/2 sin x/(sin x + cos x) dx equals:

  1. π/2
  2. π/4
  3. π
  4. 1
Show solution

Let I be the integral. By King's property (x → π/2 − x), I = ∫0π/2 cos x/(cos x + sin x) dx.

Adding: 2I = ∫0π/2 1 dx = π/2 ⇒ I = π/4.

Answer: (B)

Q3Easy

∫−11 x³ cos x dx equals:

  1. 0
  2. 2
  3. 2 cos 1
  4. 1
Show solution

x³ is odd and cos x is even, so x³ cos x is odd. Its integral over a symmetric interval is 0.

Answer: (A)

Q4Medium

The area of the region bounded by y = x and y = x² is:

  1. 1/3
  2. 1/2
  3. 1/6
  4. 2/3
Show solution

The curves meet where x = x², i.e. x = 0 and x = 1. Between them the line is above the parabola.

Area = ∫01 (x − x²) dx = 1/2 − 1/3 = 1/6.

Answer: (C)

Q5Numerical

Evaluate ∫02 [x] dx, where [·] is the greatest integer function.

Show solution

On [0, 1), [x] = 0; on [1, 2), [x] = 1. So the integral = 0 × 1 + 1 × 1 = 1.

Answer: 1

Q6Medium

∫eˣ(sin x + cos x) dx equals:

  1. eˣ cos x + C
  2. eˣ sin x + C
  3. eˣ(sin x − cos x) + C
  4. −eˣ sin x + C
Show solution

With f(x) = sin x, f′(x) = cos x, so the integral is of the form ∫eˣ(f + f′)dx = eˣ f(x) + C = eˣ sin x + C.

Answer: (B)

Q7Numerical

Evaluate ∫01 x eˣ dx.

Show solution

By parts (u = x, v = eˣ): ∫x eˣ dx = x eˣ − eˣ = eˣ(x − 1).

From 0 to 1: [e(0)] − [1 × (−1)] = 0 + 1 = 1.

Answer: 1

Q8Hard

limn→∞ [1/(n + 1) + 1/(n + 2) + … + 1/(2n)] equals:

  1. 1
  2. ln 2
  3. 1/2
  4. e
Show solution

Write the sum as Σr=1n 1/(n + r) = (1/n) Σ 1/(1 + r/n).

As a Riemann sum this is ∫01 dx/(1 + x) = ln 2.

Answer: (B)

11. Quick Revision Summary

Remember before the exam
  • Learn the standard table cold, including ∫sec x and the three a² forms.
  • Substitution for f(g(x))g′(x); by parts with ILATE; eˣ(f + f′) shortcut.
  • King's property, odd/even and periodicity crack most definite integrals.
  • Riemann sum: (1/n)Σf(r/n) → ∫01f(x)dx.
  • Area = ∫(upper − lower) between intersection points; sketch first.
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