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Definite Integration — Advanced Techniques & Leibniz Rule

Master Advanced Definite Integration for JEE Advanced. Covers Newton-Leibniz Rule for differentiating integrals with variable limits including both limits variable, FTC with L-Hopital for 0/0 limits, Differentiation Under Integral Sign DUIS Feynman technique with classic ln x problem, Integral Equations finding f(x) from functional equations, Riemann Sum limit of sum to definite integral, Estimation and Comparison of integrals with 7 fully verified JEE-level practice questions.

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This note completes the study of definite integrals with three powerful advanced tools: the Newton–Leibniz Rule (differentiating an integral whose limits depend on $x$ ), Differentiation Under the Integral Sign (DUIS) (differentiating a parameter inside the integrand), and Integral Equations (finding an unknown function that satisfies an integral condition). Together, these techniques crack a class of JEE problems that are completely inaccessible by standard methods — integrals with $\ln x$ in the denominator, limits involving $x^{2}$ or $g (x)$ , and functional equations in integral form. For JEE Advanced, at least one of these techniques appears in virtually every paper, often in 3–4 mark questions.

1. Newton–Leibniz Rule (Differentiation of Definite Integral w.r.t. a Limit) JEE Main & Advanced

When the limits of a definite integral are functions of $x$ , we differentiate using the following rule:

$\frac{d}{d x} \int_{u (x)}^{v (x)} f (t) d t = f (v (x)) \cdot v^{'} (x) - f (u (x)) \cdot u^{'} (x)$

Special Cases

Form	Derivative	Reason
$\frac{d}{d x} \int_{a}^{x} f (t) d t$	$f (x)$	Lower limit constant; $v (x) = x$ , $v^{'} (x) = 1$
$\frac{d}{d x} \int_{x}^{b} f (t) d t$	$- f (x)$	Upper limit constant; $u (x) = x$ , minus sign
$\frac{d}{d x} \int_{a}^{x^{2}} f (t) d t$	$f (x^{2}) \cdot 2 x$	Chain rule: $v (x) = x^{2}$ , $v^{'} (x) = 2 x$
$\frac{d}{d x} \int_{x}^{x^{2}} f (t) d t$	$f (x^{2}) \cdot 2 x - f (x)$	Both limits variable — apply full formula

Worked Example 1

Find $F^{'} (x)$ if $F (x) = \int_{1}^{x^{2}} \ln t d t$ .

$F^{'} (x) = \ln (x^{2}) \cdot \frac{d}{d x} (x^{2}) = 2 \ln x \cdot 2 x = 4 x \ln x$

Worked Example 2

Find $\frac{d}{d x} \int_{x}^{x^{2}} \sin t d t$ .

$= \sin (x^{2}) \cdot 2 x - \sin (x) \cdot 1 = 2 x \sin (x^{2}) - \sin x$

Application — Evaluating Limits via FTC + L'Hôpital

Many JEE limits of the form $lim_{x \to a} \frac{\int_{a}^{x} f (t) d t}{g (x)}$ are $\frac{0}{0}$ forms. Apply L'Hôpital by differentiating numerator and denominator:

$lim_{x \to a} \frac{\int_{a}^{x} f (t) d t}{g (x)} = lim_{x \to a} \frac{f (x)}{g^{'} (x)}$

Example: $lim_{x \to 0} \frac{\int_{0}^{x} t \sin t d t}{x^{2}}$

$\overset{L'H}{\to} lim_{x \to 0} \frac{x \sin x}{2 x} = lim_{x \to 0} \frac{\sin x}{2} = \frac{0}{2} = 0$

2. Differentiation Under the Integral Sign (DUIS) — Feynman's Technique JEE Advanced

If an integral contains a parameter (say $a$ ) inside the integrand, we can differentiate w.r.t. that parameter to create a simpler integral, solve it, then integrate back.

If $I (a) = \int_{c}^{d} f (x, a) d x$ , then under suitable conditions:

$I^{'} (a) = \frac{d I}{d a} = \int_{c}^{d} \frac{\partial f}{\partial a} d x$

The key idea: differentiating w.r.t. the parameter $a$ is often much simpler than computing the original integral directly.

The Classic Example: $\int_{0}^{1} \frac{x^{a} - 1}{\ln x} d x$

Let $I (a) = \int_{0}^{1} \frac{x^{a} - 1}{\ln x} d x$ .

Step 1 — Differentiate w.r.t. $a$ :

$I^{'} (a) = \int_{0}^{1} \frac{\partial}{\partial a} (\frac{x^{a} - 1}{\ln x}) d x = \int_{0}^{1} \frac{x^{a} \ln x}{\ln x} d x = \int_{0}^{1} x^{a} d x = \frac{1}{a + 1}$

Step 2 — Integrate back:

$I (a) = \int \frac{1}{a + 1} d a = \ln (a + 1) + C$

Step 3 — Determine constant using a known value:

$I (0) = \int_{0}^{1} \frac{x^{0} - 1}{\ln x} d x = \int_{0}^{1} 0 d x = 0$ . So $C = 0$ .

$∴ I (a) = \ln (a + 1)$

Now evaluate any integral of this family instantly:

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

When to Use DUIS

Look for these signals in the integrand:

$\frac{x^{a}}{\ln x}$ or $\frac{x^{a} - x^{b}}{\ln x}$ — the $\ln x$ in the denominator "cancels" on differentiation.
An integral you cannot evaluate by standard methods but could simplify by adding a parameter.
An integral family where one member is easy (e.g., $a = 0$ gives a known value).

3. Integral Equations — Finding $f (x)$ from a Functional Equation JEE Advanced

JEE Advanced frequently poses problems of the form: "Given that $f (x)$ satisfies $f (x) = g (x) + \int_{a}^{b} h (t) \cdot f (t) d t$ , find $f (x)$ ."

The key observation: $\int_{a}^{b} h (t) \cdot f (t) d t$ is a definite integral — just a number. So we call it $c$ (a constant), solve for $f (x)$ in terms of $c$ , substitute back to find $c$ .

Standard Method

Let $c = \int_{a}^{b} h (t) \cdot f (t) d t$ ( $c$ is a constant).
Then $f (x) = g (x) + c$ — express $f$ in terms of $c$ .
Substitute $f (t) = g (t) + c$ into the definition of $c$ :
Solve the resulting algebraic equation for $c$ .
Substitute back to get explicit $f (x)$ .

Worked Example

Find $f (x)$ if $f (x) = x + \int_{0}^{1} t \cdot f (t) d t$ .

Step 1: Let $c = \int_{0}^{1} t \cdot f (t) d t$ (a constant).

Step 2: Then $f (x) = x + c$ , so $f (t) = t + c$ .

Step 3: Substitute: $c = \int_{0}^{1} t (t + c) d t = \int_{0}^{1} (t^{2} + c t) d t = {[\frac{t^{3}}{3} + \frac{c t^{2}}{2}]}_{0}^{1} = \frac{1}{3} + \frac{c}{2}$

Step 4: Solve: $c - \frac{c}{2} = \frac{1}{3} \Rightarrow \frac{c}{2} = \frac{1}{3} \Rightarrow c = \frac{2}{3}$ .

Step 5: $f (x) = x + \frac{2}{3}$ .

Verification: $\int_{0}^{1} t (t + \frac{2}{3}) d t = \int_{0}^{1} (t^{2} + \frac{2 t}{3}) d t = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} = c$ ✓

4. Summation Using Integration (Limit as a Riemann Sum) JEE Advanced

This technique (introduced in Topic 3) is extended here. The master formula is:

$lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} f (\frac{r}{n}) = \int_{0}^{1} f (x) d x$

More generally: $lim_{n \to \infty} \frac{1}{n} \sum_{r = p}^{q \cdot n} f (\frac{r}{n}) = \int_{p}^{q} f (x) d x$

Step-by-Step Identification

Signal in the sum	Interpretation
$\frac{1}{n}$ multiplied overall	This is $d x$ (width of each strip)
$\frac{r}{n}$ appearing	This is $x$ (position of strip)
$r$ runs from $1$ to $n$	Limits $0$ to $1$
$r$ runs from $0$ to $n - 1$	Limits $0$ to $1$ (same integral)

Worked Example

Evaluate: $lim_{n \to \infty} (\frac{1}{\sqrt{n^{2} + 1^{2}}} + \frac{1}{\sqrt{n^{2} + 2^{2}}} + \dots + \frac{1}{\sqrt{n^{2} + n^{2}}})$

$= lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{\sqrt{n^{2} + r^{2}}} = lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{\sqrt{1 + {(\frac{r}{n})}^{2}}}$

$= \int_{0}^{1} \frac{1}{\sqrt{1 + x^{2}}} d x = [\ln (x + \sqrt{1 + x^{2}})]_{0}^{1} = \ln (1 + \sqrt{2}) - \ln 1 = \ln (1 + \sqrt{2})$

5. Estimation and Comparison of Definite Integrals JEE Advanced

When an integral cannot be computed exactly, its value can be bounded using:

Bound Property

If $m \leq f (x) \leq M$ for all $x \in [a, b]$ , then:

$m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a)$

Comparison Property

If $f (x) \geq g (x)$ for all $x \in [a, b]$ , then:

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

Example — Bounding $\int_{0}^{1} e^{- x^{2}} d x$

On $[0, 1]$ : $e^{- 1} \leq e^{- x^{2}} \leq e^{0} = 1$ . So:

$e^{- 1} \cdot 1 \leq \int_{0}^{1} e^{- x^{2}} d x \leq 1$ , i.e., $\frac{1}{e} \leq \int_{0}^{1} e^{- x^{2}} d x \leq 1$

(This integral has no closed form — the bound is the best we can do analytically.)

Advanced Techniques — Decision Framework

Problem type	Technique	Key step
$\frac{d}{d x} \int_{g (x)}^{h (x)} f (t) d t$	Newton–Leibniz	$f (h (x)) \cdot h^{'} (x) - f (g (x)) \cdot g^{'} (x)$
$lim_{x \to a} \frac{\int_{a}^{x} f (t) d t}{something \to 0}$	FTC + L'Hôpital	Differentiate integral in numerator → $f (x)$
$\int \frac{x^{a} - x^{b}}{\ln x} d x$ or $\int \frac{f (x, a)}{\ln x} d x$	DUIS	Differentiate w.r.t. parameter $a$ ; $\ln x$ cancels
$f (x) = g (x) + \int_{a}^{b} h (t) f (t) d t$	Integral equation	Set integral $= c$ ; substitute; solve for $c$ algebraically
$lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} f (\frac{r}{n})$	Riemann Sum	Identify $\frac{r}{n} = x$ , $\frac{1}{n} = d x$ , limits $0$ to $1$
Cannot evaluate exactly; need to compare/bound	Estimation	Find $m \leq f (x) \leq M$ ; bound $= m (b - a)$ to $M (b - a)$

JEE Power Tips — Avoid These Common Mistakes

Newton–Leibniz: chain rule is mandatory when the limit is not simply $x$ . $\frac{d}{d x} \int_{a}^{x^{2}} f (t) d t = f (x^{2}) \cdot 2 x$ , NOT just $f (x^{2})$ . Forgetting the $2 x$ (the derivative of $x^{2}$ ) is one of the most common JEE errors in this topic.
In DUIS, the constant of integration $C$ must be determined from a boundary value. Always find a value of $a$ where $I (a)$ can be evaluated directly (usually $a = 0$ or $a = 1$ ) and use it to pin down $C$ . Without this step, the answer is incomplete.
In integral equations: the integral $\int_{a}^{b} h (t) f (t) d t$ is a constant — it doesn't depend on $x$ . This is because $a$ , $b$ , $h$ , and $f$ are all fixed once $f$ is determined — there is no $x$ inside. Students sometimes forget this and try to differentiate it, which is wrong at this stage.
For Riemann sums: $r$ starts from 0 or 1 — both give $\int_{0}^{1}$ . When $r : 0 \to n - 1$ , the sum still converges to $\int_{0}^{1}$ . When $r : 1 \to n$ , same limit. Either way the integral is from 0 to 1. Only when $r : 1 \to 2 n$ or similar does the upper limit change.
$\frac{d}{d x} \int_{x}^{x^{2}} f (t) d t$ has TWO terms with opposite signs. Both limits are variable — upper limit gives $+ f (x^{2}) \cdot 2 x$ , lower limit gives $- f (x) \cdot 1$ . A common error is dropping the lower limit term entirely.
DUIS key identity: $\frac{\partial}{\partial a} (x^{a}) = x^{a} \ln x$ . This is why $\ln x$ in the denominator "cancels" under DUIS — differentiating $x^{a}$ introduces $\ln x$ in the numerator, which cancels with the $\ln x$ denominator. If you don't know this identity, DUIS problems become impossible.

Practice Questions (JEE Advanced Level)

Q1 (Newton–Leibniz): Find $\frac{d}{d x} \int_{0}^{x} t^{2} d t$ and evaluate it at $x = 2$ .

Answer: $x^{2}$ ; value at $x = 2$ is 4.

Explanation:

By the Fundamental Theorem of Calculus (Newton-Leibniz rule):
$\frac{d}{d x} \int_{0}^{x} t^{2} d t = x^{2}$ .

Evaluating this derivative at $x = 2$ :
$(2)^{2} = 4$ .

Verification:
Evaluate the integral first: $\int_{0}^{x} t^{2} d t = \frac{x^{3}}{3}$ .
Then differentiate: $\frac{d}{d x} (\frac{x^{3}}{3}) = \frac{3 x^{2}}{3} = x^{2}$ .

Q2 (Newton–Leibniz, variable both limits): Find $\frac{d}{d x} \int_{x}^{x^{2}} \sin t d t$ .

Answer: $2 x \sin (x^{2}) - \sin x$

Explanation:

Apply the general Leibniz rule for differentiation under the integral sign where both the upper limit $v (x) = x^{2}$ and lower limit $u (x) = x$ are functions of $x$ :

$\frac{d}{d x} \int_{u (x)}^{v (x)} f (t) d t = f (v (x)) \cdot v^{'} (x) - f (u (x)) \cdot u^{'} (x)$

Applying this to our specific integral:
$\frac{d}{d x} \int_{x}^{x^{2}} \sin t d t = \sin (x^{2}) \cdot \underset{2 x}{\underset{⏟}{\frac{d}{d x} (x^{2})}} - \sin (x) \cdot \underset{1}{\underset{⏟}{\frac{d}{d x} (x)}}$

$= 2 x \sin (x^{2}) - \sin x$ .

Q3 (DUIS): Evaluate $\int_{0}^{1} \frac{x^{3} - x^{2}}{\ln x} d x$ .

Answer: $\ln \frac{4}{3}$

Explanation:

This integral can be solved using the standard result derived from Differentiation Under the Integral Sign (Feynman's Trick):
$I (a) = \int_{0}^{1} \frac{x^{a} - 1}{\ln x} d x = \ln (a + 1)$ .

Rewrite the given integral to match this form:
$\int_{0}^{1} \frac{x^{3} - x^{2}}{\ln x} d x = \int_{0}^{1} \frac{(x^{3} - 1) - (x^{2} - 1)}{\ln x} d x$

Split the integral and apply the standard result:
$= I (3) - I (2)$
$= \ln (3 + 1) - \ln (2 + 1)$
$= \ln 4 - \ln 3$

Using logarithmic properties:
$= \ln \frac{4}{3}$ .

Q4 (FTC + L'Hôpital): Evaluate $lim_{x \to 0} \frac{1}{x^{2}} \int_{0}^{x} t \sin t d t$ .

Answer: 0

Explanation:

Rewrite the limit as a fraction:
$lim_{x \to 0} \frac{\int_{0}^{x} t \sin t d t}{x^{2}}$

As $x \to 0$ , the upper limit of the integral approaches 0, so the numerator $\to 0$ . The denominator $x^{2} \to 0$ as well. This creates an indeterminate $\frac{0}{0}$ form.

Apply L'Hôpital's rule by differentiating the numerator (using FTC) and the denominator:
$= lim_{x \to 0} \frac{\frac{d}{d x} \int_{0}^{x} t \sin t d t}{\frac{d}{d x} (x^{2})}$

$= lim_{x \to 0} \frac{x \sin x}{2 x}$

Cancel the $x$ terms:
$= lim_{x \to 0} \frac{\sin x}{2} = \frac{0}{2} = 0$ .

Q5 (Integral Equation): Find $f (x)$ if $f (x) = x + \int_{0}^{1} t \cdot f (t) d t$ .

Answer: $f (x) = x + \frac{2}{3}$

Explanation:

Step 1: Notice that the definite integral $\int_{0}^{1} t \cdot f (t) d t$ will evaluate to a constant number. Let's call this constant $c$ .
Therefore, the function takes the form: $f (x) = x + c$ .

Step 2: Substitute $f (t) = t + c$ back into the original integral definition for $c$ :
$c = \int_{0}^{1} t (t + c) d t$
$c = \int_{0}^{1} (t^{2} + c t) d t$

Step 3: Evaluate the integral and solve for $c$ :
$c = {[\frac{t^{3}}{3} + \frac{c t^{2}}{2}]}_{0}^{1}$
$c = \frac{1}{3} + \frac{c}{2}$

$c - \frac{c}{2} = \frac{1}{3}$
$\frac{c}{2} = \frac{1}{3} ⟹ c = \frac{2}{3}$

Substitute $c$ back into our function definition:
$∴ f (x) = x + \frac{2}{3}$ .

Q6 (Riemann Sum): Evaluate $lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{\sqrt{n^{2} + r^{2}}}$ .

Answer: $\ln (1 + \sqrt{2})$

Explanation:

Convert the infinite sum into a definite integral. First, factor out $n^{2}$ from the square root in the denominator:
$\sum_{r = 1}^{n} \frac{1}{\sqrt{n^{2} (1 + \frac{r^{2}}{n^{2}})}}$

Pull $\frac{1}{n}$ out of the sum to act as $d x$ :
$= \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{\sqrt{1 + {(\frac{r}{n})}^{2}}}$

As $n \to \infty$ , this Riemann sum transforms into an integral where $\frac{r}{n} \to x$ , $\frac{1}{n} \to d x$ , and the limits map from $0$ to $1$ :
$= \int_{0}^{1} \frac{d x}{\sqrt{1 + x^{2}}}$

Evaluate the standard integral:
$= [\ln (x + \sqrt{1 + x^{2}})]_{0}^{1}$
$= \ln (1 + \sqrt{1 + 1}) - \ln (0 + \sqrt{1 + 0})$
$= \ln (1 + \sqrt{2}) - \ln (1)$
$= \ln (1 + \sqrt{2})$ .

Q7 (Newton–Leibniz + FTC): If $F (x) = \int_{1}^{x^{2}} \ln t d t$ , find $F^{'} (x)$ .

Answer: $4 x \ln x$

Explanation:

Apply the Newton-Leibniz rule:
$F^{'} (x) = \ln (x^{2}) \cdot \frac{d}{d x} (x^{2}) - \ln (1) \cdot \frac{d}{d x} (1)$

$F^{'} (x) = \ln (x^{2}) \cdot (2 x) - 0$

Using the logarithm power rule ( $\ln (x^{2}) = 2 \ln x$ for $x > 0$ ):
$F^{'} (x) = (2 \ln x) \cdot (2 x) = 4 x \ln x$ .

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Definite Integration — Advanced Techniques & Leibniz Rule

1. Newton–Leibniz Rule (Differentiation of Definite Integral w.r.t. a Limit) JEE Main & Advanced

Special Cases

Worked Example 1

Worked Example 2

Application — Evaluating Limits via FTC + L'Hôpital

2. Differentiation Under the Integral Sign (DUIS) — Feynman's Technique JEE Advanced

The Classic Example: $\int_{0}^{1} \frac{x^{a} - 1}{\ln x} d x$

When to Use DUIS

3. Integral Equations — Finding $f (x)$ from a Functional Equation JEE Advanced

Standard Method

Worked Example

4. Summation Using Integration (Limit as a Riemann Sum) JEE Advanced

Step-by-Step Identification

Worked Example

5. Estimation and Comparison of Definite Integrals JEE Advanced

Bound Property

Comparison Property

Example — Bounding $\int_{0}^{1} e^{- x^{2}} d x$

Advanced Techniques — Decision Framework

Practice Questions (JEE Advanced Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

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Subject Mastery

Definite Integration — Advanced Techniques & Leibniz Rule

1. Newton–Leibniz Rule (Differentiation of Definite Integral w.r.t. a Limit) JEE Main & Advanced

Special Cases

Worked Example 1

Worked Example 2

Application — Evaluating Limits via FTC + L'Hôpital

2. Differentiation Under the Integral Sign (DUIS) — Feynman's Technique JEE Advanced

The Classic Example: ∫01xa−1ln⁡xdx

When to Use DUIS

3. Integral Equations — Finding f(x) from a Functional Equation JEE Advanced

Standard Method

Worked Example

4. Summation Using Integration (Limit as a Riemann Sum) JEE Advanced

Step-by-Step Identification

Worked Example

5. Estimation and Comparison of Definite Integrals JEE Advanced

Bound Property

Comparison Property

Example — Bounding ∫01e−x2dx

Advanced Techniques — Decision Framework

Practice Questions (JEE Advanced Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

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We Value Your Privacy

The Classic Example: $\int_{0}^{1} \frac{x^{a} - 1}{\ln x} d x$

3. Integral Equations — Finding $f (x)$ from a Functional Equation JEE Advanced

Example — Bounding $\int_{0}^{1} e^{- x^{2}} d x$