What if f(x) is discontinuous inside the interval of integration?

Newton-Leibnitz requires continuity. If f has a discontinuity at c inside (a, b), split the integral at c and evaluate each piece separately. If the discontinuity is unbounded, the integral becomes improper.

Definite Integration And Fundamental Theorem

Q: What is the difference between a definite integral and area?

A definite integral gives the algebraic sum of signed areas (positive above the x-axis, negative below). True area is always non-negative and is computed using the absolute value of f(x). They agree only when f(x) has constant sign on the interval.

Q: Why did my substitution give a wrong answer with a negative sign?

A substitution x = phi(t) is valid only if phi(t) is continuous on the interval of integration. Using x = 1/t across an interval containing t = 0 invalidates the substitution and can flip the sign of the answer.

Q: Can a definite integral be zero?

Yes. If positive and negative signed areas cancel exactly, the integral is zero. For example, the integral of x cubed from -1 to 1 is zero because x cubed is odd, so the two signed areas cancel.

Q: How do I handle a piecewise function inside a definite integral?

Split the interval at every point where the definition or the sign of a modulus changes, evaluate each piece using the appropriate formula, and add the results.

Q: Which antiderivative do I choose for Newton-Leibnitz?

Any antiderivative works. Two antiderivatives of f differ by a constant, and that constant cancels in F(b) - F(a). Drop the +C when applying the Fundamental Theorem.

MathsIntegralsFor JEE aspirants

The definite integral $\int_{a}^{b} f (x) d x$ evaluates the accumulated signed area between the curve $y = f (x)$ and the x-axis from $x = a$ to $x = b$ . The Fundamental Theorem of Calculus (also called the Newton-Leibnitz formula) gives the shortcut: if $F (x)$ is any antiderivative of $f (x)$ on $[a, b]$ , then $\int_{a}^{b} f (x) d x = F (b) - F (a)$ . This concept is the gateway to every JEE definite-integration problem, connecting antiderivatives to signed areas and setting up all further techniques.

Key Formulas - Quick Reference

Newton-Leibnitz (FTC): $\int_{a}^{b} f (x) d x = F (b) - F (a)$ , where $F^{'} (x) = f (x)$
Signed area: $\int_{a}^{b} f (x) d x =$ (area above x-axis) $-$ (area below x-axis)
True area: Area $= \int_{a}^{b} ∣ f (x) ∣ d x$ (always non-negative)
Reversal: $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$
Continuity requirement: Substitution $x = ϕ (t)$ is valid only if $ϕ (t)$ is continuous on the interval of integration

1. Geometrical Interpretation of the Definite Integral

If $f (x) > 0$ for all $x \in [a, b]$ , then $\int_{a}^{b} f (x) d x$ is numerically equal to the area bounded by the curve $y = f (x)$ , the x-axis, and the vertical lines $x = a$ and $x = b$ .

In general, $\int_{a}^{b} f (x) d x$ represents the algebraic sum of the areas bounded by the curve $y = f (x)$ , the x-axis, and the lines $x = a$ and $x = b$ . Areas above the x-axis are taken with a plus sign; areas below the x-axis are taken with a minus sign.

Figure 1: Geometrical meaning of

\int_{a}^{b} f (x) d x

. Regions above the x-axis contribute positively; regions below contribute negatively.

Important distinction: If someone asks for the area of

y = f (x)

between

a

and

b

, the answer is

\int_{a}^{b} ∣ f (x) ∣ d x

, not

\int_{a}^{b} f (x) d x

. The two agree only when

f (x)

has constant sign on

[a, b]

Illustrative example: $y = x^{3}$ on $[- 1, 1]$

Consider the area bounded by $y = x^{3}$ between $x = - 1$ and $x = 1$ . Because $x^{3} < 0$ on $[- 1, 0]$ and $x^{3} > 0$ on $[0, 1]$ :

$Area = \int_{- 1}^{0} (- x^{3}) d x + \int_{0}^{1} x^{3} d x = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

Equivalently, using the absolute-value formulation:

$Area = \int_{- 1}^{1} ∣ x^{3} ∣ d x = 2 \int_{0}^{1} x^{3} d x = 2 \cdot \frac{1}{4} = \frac{1}{2}$

But if we compute the definite integral directly:

$\int_{- 1}^{1} x^{3} d x = [\frac{x ^{4}}{4}]_{- 1}^{1} = \frac{1}{4} - \frac{1}{4} = 0$

This value of $0$ is the algebraic sum, not the area. Always separate the two ideas in your mind.

2. Fundamental Theorem of Calculus (Newton-Leibnitz Formula)

Statement: If

f (x)

is a continuous function on

[a, b]

and

F (x)

is any antiderivative of

f (x)

[a, b]

(that is,

F^{'} (x) = f (x)

for all

x \in [a, b]

), then

\int_{a}^{b} f (x) d x = F (b) - F (a)

The function $F (x)$ is called the integral of $f (x)$ ; $a$ and $b$ are the lower and upper limits of integration respectively.

Note: The theorem requires

f

to be continuous on

[a, b]

. If

f

has a discontinuity inside

[a, b]

, split the interval at the discontinuity and evaluate each piece separately.

Step-by-step procedure

Find an antiderivative $F (x)$ of $f (x)$ using standard integration techniques.
Evaluate $F$ at the upper limit: compute $F (b)$ .
Evaluate $F$ at the lower limit: compute $F (a)$ .
Subtract: $\int_{a}^{b} f (x) d x = F (b) - F (a)$ .

Continuity of the substitution

When using a substitution $x = ϕ (t)$ to evaluate a definite integral, the substitution must be continuous on the interval of integration. If $ϕ (t)$ is discontinuous at any point in the interval, the substitution is invalid and can produce a wrong (often negative or absurd) answer.

Solved Example 1

Evaluate

\int_{1}^{2} \frac{d x}{( x + 1 ) ( x + 2 )}

Solution:

Using partial fractions: $\frac{1}{( x + 1 ) ( x + 2 )} = \frac{1}{x + 1} - \frac{1}{x + 2}$

$\int_{1}^{2} \frac{d x}{( x + 1 ) ( x + 2 )} = [ln (x + 1) - ln (x + 2)]_{1}^{2}$

$= (ln 3 - ln 4) - (ln 2 - ln 3) = 2 ln 3 - ln 4 - ln 2 = ln \frac{9}{8}$

Solved Example 2

Evaluate

\int_{2}^{5} f (x) d x

where

f (x) = x + 3

for

x < 3

, and

f (x) = 3 x^{2} + 1

for

x \geq 3

Solution:

Because $f$ is defined piecewise, split at $x = 3$ :

$\int_{2}^{5} f (x) d x = \int_{2}^{3} (x + 3) d x + \int_{3}^{5} (3 x^{2} + 1) d x$

$= [\frac{x ^{2}}{2} + 3 x]_{2}^{3} + [x^{3} + x]_{3}^{5}$

$= (\frac{9}{2} + 9) - (2 + 6) + (125 + 5) - (27 + 3)$

$= \frac{9 - 4}{2} + 3 + 5^{3} - 3^{3} + 5 - 3 = \frac{5}{2} + 3 + 98 + 2 = \frac{211}{2}$

Solved Example 3

Evaluate

\int_{2}^{8} ∣ x - 5∣ d x

Solution:

The expression $∣ x - 5∣$ changes sign at $x = 5$ . Split the interval:

$\int_{2}^{8} ∣ x - 5∣ d x = \int_{2}^{5} (5 - x) d x + \int_{5}^{8} (x - 5) d x$

$= [5 x - \frac{x ^{2}}{2}]_{2}^{5} + [\frac{x ^{2}}{2} - 5 x]_{5}^{8}$

$= (25 - \frac{25}{2}) - (10 - 2) + (32 - 40) - (\frac{25}{2} - 25)$

$= \frac{25}{2} - 8 - 8 + \frac{25}{2} = 25 - 16 = 9$

Solved Example 4

Evaluate

I = \int_{- 2}^{2} \frac{d x}{4 + x ^{2}}

directly, then check by the substitution

x = 1/ t

. Explain why the two methods disagree.

Solution:

Direct evaluation:

$I = \int_{- 2}^{2} \frac{d x}{4 + x ^{2}} = [\frac{1}{2} tan^{- 1} (\frac{x}{2})]_{- 2}^{2}$

$= \frac{1}{2} [tan^{- 1} (1) - tan^{- 1} (- 1)] = \frac{1}{2} [\frac{π}{4} + \frac{π}{4}] = \frac{π}{4}$

Substitution $x = 1/ t$ : Then $d x = - d t / t^{2}$ . Limits: when $x = - 2$ , $t = - 1/2$ ; when $x = 2$ , $t = 1/2$ .

$I = \int_{- 1/2}^{1/2} \frac{- \frac{1}{t ^{2}} d t}{4 + \frac{1}{t ^{2}}} = - \int_{- 1/2}^{1/2} \frac{d t}{4 t ^{2} + 1} = - [\frac{1}{2} tan^{- 1} (2 t)]_{- 1/2}^{1/2}$

$= - \frac{1}{2} tan^{- 1} (1) - (- \frac{1}{2} tan^{- 1} (- 1)) = - \frac{π}{8} - \frac{π}{8} = - \frac{π}{4}$

Why the disagreement? The integrand $\frac{1}{4 + x ^{2}} > 0$ everywhere, so the definite integral cannot be negative. The correct answer is $π /4$ . The substitution $x = 1/ t$ is discontinuous at $t = 0$ , which lies inside the transformed interval $[- 1/2, 1/2]$ , so the substitution is invalid. This confirms the earlier note: the substitution must be continuous on the entire interval of integration.

Solved Example 5

Let

α = \int_{0}^{\infty} \frac{d x}{x ^{4} + 7 x ^{2} + 1}

and

β = \int_{0}^{\infty} \frac{x ^{2} d x}{x ^{4} + 7 x ^{2} + 1}

. Prove that

α = β

Solution:

In $α$ , substitute $x = 1/ t$ , so $d x = - d t / t^{2}$ . The limits swap: as $x : 0 \to \infty$ , $t : \infty \to 0$ .

$α = \int_{\infty}^{0} \frac{- \frac{1}{t ^{2}} d t}{\frac{1}{t ^{4}} + \frac{7}{t ^{2}} + 1} = \int_{0}^{\infty} \frac{\frac{1}{t ^{2}} d t}{\frac{1 + 7 t ^{2} + t ^{4}}{t ^{4}}} = \int_{0}^{\infty} \frac{t ^{2} d t}{t ^{4} + 7 t ^{2} + 1}$

The variable of integration is a dummy, so this equals $β$ . Hence $α = β$ .

Note on validity: Here the substitution $x = 1/ t$ is fine because the interval $(0, \infty)$ does not contain $t = 0$ ; the discontinuity of $1/ t$ lies outside the domain.

Solved Example 6

Evaluate

\int_{0}^{π /4} sec^{2} x d x

Solution:

The antiderivative of $sec^{2} x$ is $tan x$ . Applying Newton-Leibnitz:

$\int_{0}^{π /4} sec^{2} x d x = [tan x]_{0}^{π /4} = tan \frac{π}{4} - tan 0 = 1 - 0 = 1$

Common Mistakes to Avoid

Watch out

Confusing the definite integral with the area. $\int_{a}^{b} f (x) d x$ can be zero or negative; area is always non-negative. When asked for area, use $\int_{a}^{b} ∣ f (x) ∣ d x$ .
Forgetting to split a piecewise function or modulus at the point where the definition (or sign) changes.
Applying substitution without checking that it is continuous on the interval. A substitution with a discontinuity inside the interval can flip the sign of the answer.
Forgetting to change the limits after substitution. If you substitute $u = g (x)$ , the new limits are $g (a)$ and $g (b)$ , not $a$ and $b$ .
Writing $\int_{a}^{b} f (x) d x = - \int_{a}^{b} f (x) d x$ instead of $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$ . The reversal swaps the limits, not the sign of the integrand.
Applying Newton-Leibnitz when $f$ is discontinuous inside $[a, b]$ . The theorem requires continuity.

Frequently Asked Questions

Q1. What is the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus (Newton-Leibnitz formula) states that if $f (x)$ is continuous on $[a, b]$ and $F (x)$ is any antiderivative of $f (x)$ , then $\int_{a}^{b} f (x) d x = F (b) - F (a)$ . It connects differentiation and integration: the definite integral is the difference of antiderivative values at the endpoints.

Q2. What is the difference between a definite integral and area?

A definite integral $\int_{a}^{b} f (x) d x$ gives the algebraic sum of signed areas (positive above the x-axis, negative below). True area is always non-negative and is computed as $\int_{a}^{b} ∣ f (x) ∣ d x$ . They agree only when $f (x)$ has constant sign on $[a, b]$ .

Q3. Why did my substitution give a wrong answer with a negative sign?

A substitution $x = ϕ (t)$ is valid only if $ϕ (t)$ is continuous on the interval of integration. A common error is using $x = 1/ t$ across an interval containing $t = 0$ ; the discontinuity invalidates the substitution and can flip the sign. Always check continuity of the substitution before applying it.

Q4. Can a definite integral be zero?

Yes. If the positive and negative signed areas cancel exactly, the integral is zero. For instance, $\int_{- 1}^{1} x^{3} d x = 0$ because $x^{3}$ is odd, so the area on $[- 1, 0]$ cancels the area on $[0, 1]$ . This does not mean the region has zero area, only that the signed areas cancel.

Q5. How do I handle a piecewise function inside a definite integral?

Split the interval at every point where the definition (or the sign of a modulus) changes, evaluate each piece using the appropriate formula, and add the results. For $\int_{a}^{b} ∣ x - c ∣ d x$ with $a < c < b$ , split at $x = c$ .

Q6. Which antiderivative do I choose for Newton-Leibnitz?

Any antiderivative works. Two antiderivatives of $f$ differ by a constant, and that constant cancels in $F (b) - F (a)$ . For simplicity, drop the $+ C$ when applying the Fundamental Theorem.

Q7. What if $f (x)$ is discontinuous inside $[a, b]$ ?

Newton-Leibnitz requires continuity. If $f$ has a discontinuity at $c \in (a, b)$ , split the integral as $\int_{a}^{c} + \int_{c}^{b}$ and evaluate each piece. If the discontinuity is unbounded (an infinite jump), the integral becomes improper and needs a limit-based definition.

Previous year questions on Definite Integration And Fundamental Theorem

16 questions from past papers, each with a step-by-step solution.

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1. Geometrical Interpretation of the Definite Integral

Illustrative example: y=x3 on [−1,1]

2. Fundamental Theorem of Calculus (Newton-Leibnitz Formula)

Step-by-step procedure

Continuity of the substitution

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the Fundamental Theorem of Calculus?

Q2. What is the difference between a definite integral and area?

Q3. Why did my substitution give a wrong answer with a negative sign?

Q4. Can a definite integral be zero?

Q5. How do I handle a piecewise function inside a definite integral?

Q6. Which antiderivative do I choose for Newton-Leibnitz?

Q7. What if f(x) is discontinuous inside [a,b]?

Previous year questions on Definite Integration And Fundamental Theorem

Ready to master Integrals?

Illustrative example: $y = x^{3}$ on $[- 1, 1]$

Q7. What if $f (x)$ is discontinuous inside $[a, b]$ ?