What is Definite Integration?

Definite integration refers to the process of calculating the integral of a function between specified limits, resulting in a numerical value representing the area under the curve.

What are the common rules for definite integrals?

ome common rules include: Fundamental Theorem of Calculus Linearity rule Substitution rule Integration by parts

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Indefinite Integration

Indefinite Integration, also known as anti-differentiation, is a fundamental concept in calculus. It involves finding a function...

Double Integral

Double integration is a mathematical process used to compute the integral of a function over a two-dimensional region. It involves...

Integrals of Particular Functions

Integrals of Particular Functions have significant real-life applications, such as finding areas between curves, volumes, average v...

Important Integration Formulas

This blog covers a list of all integral formulas that are essential for JEE, along with their examples to help you understand their applications better.

Complex Numbers

A Complex Number is a Mathematical entity that combines a real number and an imaginary number. It is expressed..

Chain Rule Questions

The Chain Rule is a fundamental concept in calculus that allows us to find the derivative of composite functions. By breaking...

Coplanar Vectors

Vectors are said to be coplanar if they lie within the same plane. This means that any...

Definite Integration

The definite integral has a unique value. A definite integral is denoted by $\int_{a}^{b} f (x) d x$ , where a is called the lower limit of the integral and b is called the upper limit of the integral.

1.0Definite Integration Definition

A definite integral is denoted by $\int_{a}^{b} f (x) d x$ which represents the algebraic area bounded by the curve y = f(x), the ordinates x = a, x = b and the x-axis.

2.0Fundamental Theorem of Calculus

P–1 : If f is continuous on [a, b] then the function g defined by g(x) = $\int_{a}^{x} f (t) d t$ , a ≤ x ≤ b is continuous on [a, b] and differentiable on (a, b) and g'(x) = f(x)

P–1 : If f is continuous on [a, b] then $\int_{a}^{b} f (x) d x$ = F(b) – F(a) where F is any antiderivative of F, such that F' = f

3.0Properties of Definite Integrals

P–1 : $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t$ (change of variable does not change value of integral)

P–2 : $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$

P–3 : $\int_{a}^{b} f (x) = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$

P–4 : $\int_{- a}^{a} f (x) d x = \int_{0}^{a} (f (x) + f (- x)) d x = [0 2 \int_{0}^{a} f (x) d x if f (x) is odd if f (x) is even$

P–5: $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x or \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ (King)

P–6: $\int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x =$

$[0 2 \int_{0}^{a} f (x) d x if f (2 a - x) = - f (x) if f (2 a - x) = f (x)$ (Queen)

4.0Walli’s Theorem

(a) $\int_{0}^{π /2} sin^{n} x d x = \int_{0}^{π /2} cos^{n} x d x = \frac{( n - 1 ) ( n - 3 ) \dots .. ( 1 or 2 )}{n ( n - 2 ) \dots .. ( 1 or 2 )} K$

$where K = {π /2 1 if n is even if n is odd$

(b) $\int_{0}^{π /2} sin^{n} x \cdot cos^{m} x d x = \frac{[( n - 1 ) ( n - 3 ) ( n - 5 ) \dots .1 or 2 ] [( m - 1 ) ( m - 3 ) \dots . \dots or 2 ]}{( m + n ) ( m + n - 2 ) ( m + n - 4 ) \dots .1 or 2} K Where K = {\frac{π}{2} 1 if both m and n are even (m, n \in N) otherwise$

Related Video:

5.0Derivative of Antiderivatives(Newton-Leibnitz Theorem)

$If F (x) = \int_{g (x)}^{h (x)} f (t) d t, then \frac{d F ( x )}{d x} = h^{'} (x) f (h (x)) - g^{'} (x) f (g (x))$

$Proof: Let P (t) = \int f (t) d t \Rightarrow F (x) = \int_{g (x)}^{h (x)} f (t) d t = P (h (x)) - P (g (x))$

$\Rightarrow \frac{d F ( x )}{d x} = P^{'} (h (x)) h^{'} (x) - P^{'} (g (x)) g^{'} (x) = f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x)$

$If F (x) =, then = h^{'} (x) f (h (x)) - g^{'} (x) f (g (x))$

$Proof : Let P (t) = \int f (t) d t \Rightarrow F (x) == P (h (x)) - P (g (x))$

$\Rightarrow \frac{d F ( x )}{d x} = P^{'} (h (x)) h^{'} (x) - P^{'} (g (x)) g^{'} (x) = f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x)$

6.0Definite Integration as the limit of Sum

A definite integral can be evaluated using Riemann sums, which approximate the area under the curve by summing the areas of rectangles

$\int_{a}^{b} f (x) d x = lim_{n \to \infty} \sum_{i = 1}^{n} f (x_{i}) Δ x$

where

$Δ x = \frac{b - a}{n}$

and x_i is a sample point in the i^th subinterval.

7.0Estimation of Definite Integral and General Inequality

f(x) ≤ g(x) ≤ h(x) in [a, b] then $\int_{a}^{b} f (x) d x \leq \int_{a}^{b} g (x) d x \leq \int_{a}^{b} h (x) d x$
If M and M are respectively the least and greatest value of f(x) in [a, b] then m (b – a) ≤ $\int_{a}^{b} f (x) d x \leq M (b - a)$

Related Video:

8.0Definite Integration Questions

Example 1: $\int_{0}^{l n 2} \frac{e ^{x}}{1 + e ^{x}} d x$

Solution:

$1 + e^{x} = t x = 0 \Rightarrow t = 2$

$e^{x} d x = d t x = ln 2 \Rightarrow t = 1 + 2 = 3$

$\int_{2}^{3} \frac{d t}{t} = (ln ∣ t ∣)_{2}^{3} = ln (\frac{3}{2})$

Example 2: $\int_{α}^{β} \frac{d x}{( x - α ) ( β - x )} β > α$

Solution:

Put $x - α = t$

$\int \frac{2 t d t}{t ( β - ( t ^{2} + α )} \Rightarrow 2 \int \frac{d t}{( β - α ) - t ^{2}}$

$\Rightarrow 2 sin^{- 1} (\frac{t}{β - α}) \Rightarrow [2 sin^{- 1} (\frac{x - α}{β - α})]_{α}^{β}$

$\Rightarrow 2 (\frac{π}{2} - 0) \Rightarrow π$

Example 3: $\int_{- π /2}^{π /2} sin^{4} x cos^{6} x d x =$

(A) $\frac{3 π}{64}$

(B) $\frac{3 π}{572}$

(C) $\frac{3 π}{256}$

(D) $\frac{3 π}{128}$

Ans. (C)

Solution:

$I = \int_{- π /2}^{π /2} sin^{4} x cos^{6} x d x = 2 \int_{0}^{π /2} sin^{4} x cos^{6} x \cdot d x = 2 \frac{( 3.1 ) ( 5 \cdot 3 \cdot 1 )}{10.8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{π}{2} = \frac{3 π}{256}$

Example 4: $\int_{0}^{1} \frac{t a n ^{- 1} x}{x} d x = λ \int_{0}^{π /2} \frac{θ}{s i n θ} d θ$ , then find λ.

Ans. 1/2

Solution:

$L.H.S. \int \frac{t a n ^{- 1} x}{x} d x$

Put

$= \int_{0}^{π /4} \frac{t}{t a n t} \cdot sec^{2} t d t = \int_{0}^{π /4} \frac{t}{s i n t c o s t} d t = \int_{0}^{π /4} \frac{2 t}{s i n 2 t} d t$

2t = θ

2dt= dθ

$= \frac{1}{2} \int_{0}^{π /2} \frac{θ}{s i n θ} d θ$

∴ $λ = \frac{1}{2}$

Example 5: Evaluate $\int_{2}^{8} ∣ x - 5∣ d x$ .

Solution:

$\int_{2}^{8} ∣ x - 5∣ d x = \int_{2}^{5} (- x + 5) d x + \int_{5}^{8} (x - 5) d x = 9$

Example 6: $Evaluate \int_{- 1}^{1} \frac{3 ^{x} + 3 ^{- x}}{1 + 3 ^{x}} d x$

Solution:

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

$= \int_{0}^{1} (3^{x} + 3^{- x}) d x = (\frac{3 ^{x}}{l n 3} - \frac{3 ^{- x}}{l n 3})_{0}^{1} = (\frac{3}{l n 3} - \frac{3 ^{- 1}}{l n 3}) - (\frac{1}{l n 3} - \frac{1}{l n 3}) = \frac{1}{l n 3} [3 - \frac{1}{3}] = \frac{8}{3 l n 3}$

Example 7: $Prove that \int_{0}^{π /2} lo g (sin x) d x = \int_{0}^{π /2} lo g (cos x) d x = - \frac{π}{2} lo g 2$

Solution:

Let $I = \int_{0}^{π /2} lo g (sin x) d x$ ...(i)

then $I = \int_{0}^{π /2} lo g sin (\frac{π}{2} - x) d x = \int_{0}^{π /2} lo g (cos x) d x$ ...(ii)

adding (i) and (ii), we get

$2 I = \int_{0}^{π /2} lo g sin x d x + \int_{0}^{π /2} lo g cos x d x = \int_{0}^{π /2} (lo g sin x + lo g cos x) d x$

$\Rightarrow 2 I = \int_{0}^{π /2} lo g (sin x cos x) d x = \int_{0}^{π /2} lo g (\frac{2 s i n x c o s x}{2}) d x$

$= \int_{0}^{π /2} lo g (\frac{s i n 2 x}{2}) d x = \int_{0}^{π /2} lo g (sin 2 x) d x - \int_{0}^{π /2} (lo g 2) d x = \int_{0}^{π /2} lo g sin 2 x \cdot d x - (lo g 2) (x)_{0}^{π /2}$

$\Rightarrow 2 I = \int_{0}^{π /2} lo g (sin 2 x) d x - \frac{π}{2} lo g 2$ ...(iii)

Let

$I_{1} = \int_{0}^{π /2} lo g (sin 2 x) d x$ ,

putting 2 x=t, we get

$I_{1} = \int_{0}^{π} lo g (sin t) \frac{d t}{2} = \frac{1}{2} \int_{0}^{π} lo g (sin t) d t = \frac{1}{2} \cdot 2 \int_{0}^{π /2} lo g (sin t) d t$

$I_{1} = \int_{0}^{π /2} lo g (sin x) d x$

∴ (iii) becomes; $2 I = I - \frac{π}{2} lo g 2$

Hence $\int_{0}^{π /2} lo g sin x d x = - \frac{π}{2} lo g 2$

Example 8: Find the slope of the tangent to the curve $y = \int_{x}^{x^{2}} cos^{- 1} t^{2} d t at x = \frac{1}{4 2}$

Solution:

Given curve is

$y = \int_{x}^{x^{2}} cos^{- 1} t^{2} d t; \frac{d y}{d x} = \frac{d}{d x} \int_{x}^{x^{2}} cos^{- 1} t^{2} d t$

using Leibnitz theorem, $\frac{d y}{d x} = 2 x cos^{- 1} x^{4} - cos^{- 1} x^{2}$

$(\frac{d y}{d x})_{x = \frac{1}{4 2}} = \frac{2}{2 ^{1/4}} cos^{- 1} \frac{1}{2} - cos^{- 1} \frac{1}{2} = 2^{3/4} \frac{π}{3} - \frac{π}{4} = (\frac{4 8}{3} - \frac{1}{4}) π$

9.0Practice Problems on Definite Integration

a. If $f (x) = ∣ x ∣ + ∣ x - 1∣ + ∣ x - 2∣, x \in R then \int_{0}^{3} f (x) dx =$

b. Evaluate: $\int_{0}^{1} \frac{x ^{2} - 1}{1 + x ^{2} + x ^{4}} d x$

c. Evaluate: $\int_{0}^{π /2} \frac{d x}{1 + c o s θ \cdot c o s x} θ \in (0, π)$

d. The value of the integral $\int_{- 1}^{1} lo g (x + x^{2} + 1) d x$ is:

e. Evaluate: $\int_{- 1}^{1} \frac{( 2 x ^{332} + x ^{998} + 4 x ^{1668} \cdot s i n x ^{691} )}{1 + x ^{666}} d x$

Answers:

a. $\frac{19}{2}$

b. $\frac{1}{2} ln (\frac{1}{3})$

c. $\frac{θ}{s i n θ}$

d. 0

e. $\frac{2}{333} [\frac{π}{4} + 1] = \frac{π + 4}{666}$

10.0Solved Questions on Definite Integration

1. How do you define a definite integral?

Ans: A definite integral is defined as the integral of a function f(x) from a to b, denoted as $\int_{a}^{b} f (x) d x$ , where 'a' and 'b' are the limits of integration.

2. How do you perform definite integration by parts?

Ans: Integration by parts for definite integrals is given by: $\int_{a}^{b} u d v = uv ∣_{a}^{b} - \int_{a}^{b} v d u$ where u and v are differentiable functions.

3. How do you evaluate a definite integral using the limit of a sum?

Ans: A definite integral can be evaluated using Riemann sums, which approximate the area under the curve by summing the areas of rectangles:

$\int_{a}^{b} f (x) d x = lim_{n \to \infty} \sum_{i = 1}^{n} f (x_{i}^{*}) Δ x$ , where $Δ x = \frac{b - a}{n} and x_{i}^{*}$ is a sample point in the i^th subinterval.

Join ALLEN!

(Session 2026 - 27)

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I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Indefinite Integration

Indefinite Integration, also known as anti-differentiation, is a fundamental concept in calculus. It involves finding a function...

Double Integral

Double integration is a mathematical process used to compute the integral of a function over a two-dimensional region. It involves...

Integrals of Particular Functions

Integrals of Particular Functions have significant real-life applications, such as finding areas between curves, volumes, average v...

Important Integration Formulas

This blog covers a list of all integral formulas that are essential for JEE, along with their examples to help you understand their applications better.

Complex Numbers

A Complex Number is a Mathematical entity that combines a real number and an imaginary number. It is expressed..

Chain Rule Questions

The Chain Rule is a fundamental concept in calculus that allows us to find the derivative of composite functions. By breaking...

Coplanar Vectors

Vectors are said to be coplanar if they lie within the same plane. This means that any...

Definite Integration

The definite integral has a unique value. A definite integral is denoted by $\int_{a}^{b} f (x) d x$ , where a is called the lower limit of the integral and b is called the upper limit of the integral.

1.0Definite Integration Definition

A definite integral is denoted by $\int_{a}^{b} f (x) d x$ which represents the algebraic area bounded by the curve y = f(x), the ordinates x = a, x = b and the x-axis.

2.0Fundamental Theorem of Calculus

P–1 : If f is continuous on [a, b] then the function g defined by g(x) = $\int_{a}^{x} f (t) d t$ , a ≤ x ≤ b is continuous on [a, b] and differentiable on (a, b) and g'(x) = f(x)

P–1 : If f is continuous on [a, b] then $\int_{a}^{b} f (x) d x$ = F(b) – F(a) where F is any antiderivative of F, such that F' = f

3.0Properties of Definite Integrals

P–1 : $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t$ (change of variable does not change value of integral)

P–2 : $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$

P–3 : $\int_{a}^{b} f (x) = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$

P–4 : $\int_{- a}^{a} f (x) d x = \int_{0}^{a} (f (x) + f (- x)) d x = [0 2 \int_{0}^{a} f (x) d x if f (x) is odd if f (x) is even$

P–5: $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x or \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ (King)

P–6: $\int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x =$

$[0 2 \int_{0}^{a} f (x) d x if f (2 a - x) = - f (x) if f (2 a - x) = f (x)$ (Queen)

4.0Walli’s Theorem

(a) $\int_{0}^{π /2} sin^{n} x d x = \int_{0}^{π /2} cos^{n} x d x = \frac{( n - 1 ) ( n - 3 ) \dots .. ( 1 or 2 )}{n ( n - 2 ) \dots .. ( 1 or 2 )} K$

$where K = {π /2 1 if n is even if n is odd$

(b) $\int_{0}^{π /2} sin^{n} x \cdot cos^{m} x d x = \frac{[( n - 1 ) ( n - 3 ) ( n - 5 ) \dots .1 or 2 ] [( m - 1 ) ( m - 3 ) \dots . \dots or 2 ]}{( m + n ) ( m + n - 2 ) ( m + n - 4 ) \dots .1 or 2} K Where K = {\frac{π}{2} 1 if both m and n are even (m, n \in N) otherwise$

Related Video:

5.0Derivative of Antiderivatives(Newton-Leibnitz Theorem)

$If F (x) = \int_{g (x)}^{h (x)} f (t) d t, then \frac{d F ( x )}{d x} = h^{'} (x) f (h (x)) - g^{'} (x) f (g (x))$

$Proof: Let P (t) = \int f (t) d t \Rightarrow F (x) = \int_{g (x)}^{h (x)} f (t) d t = P (h (x)) - P (g (x))$

$\Rightarrow \frac{d F ( x )}{d x} = P^{'} (h (x)) h^{'} (x) - P^{'} (g (x)) g^{'} (x) = f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x)$

$If F (x) =, then = h^{'} (x) f (h (x)) - g^{'} (x) f (g (x))$

$Proof : Let P (t) = \int f (t) d t \Rightarrow F (x) == P (h (x)) - P (g (x))$

$\Rightarrow \frac{d F ( x )}{d x} = P^{'} (h (x)) h^{'} (x) - P^{'} (g (x)) g^{'} (x) = f (h (x)) h^{'} (x) - f (g (x)) g^{'} (x)$

6.0Definite Integration as the limit of Sum

A definite integral can be evaluated using Riemann sums, which approximate the area under the curve by summing the areas of rectangles

$\int_{a}^{b} f (x) d x = lim_{n \to \infty} \sum_{i = 1}^{n} f (x_{i}) Δ x$

where

$Δ x = \frac{b - a}{n}$

and x_i is a sample point in the i^th subinterval.

7.0Estimation of Definite Integral and General Inequality

f(x) ≤ g(x) ≤ h(x) in [a, b] then $\int_{a}^{b} f (x) d x \leq \int_{a}^{b} g (x) d x \leq \int_{a}^{b} h (x) d x$
If M and M are respectively the least and greatest value of f(x) in [a, b] then m (b – a) ≤ $\int_{a}^{b} f (x) d x \leq M (b - a)$

Related Video:

8.0Definite Integration Questions

Example 1: $\int_{0}^{l n 2} \frac{e ^{x}}{1 + e ^{x}} d x$

Solution:

$1 + e^{x} = t x = 0 \Rightarrow t = 2$

$e^{x} d x = d t x = ln 2 \Rightarrow t = 1 + 2 = 3$

$\int_{2}^{3} \frac{d t}{t} = (ln ∣ t ∣)_{2}^{3} = ln (\frac{3}{2})$

Example 2: $\int_{α}^{β} \frac{d x}{( x - α ) ( β - x )} β > α$

Solution:

Put $x - α = t$

$\int \frac{2 t d t}{t ( β - ( t ^{2} + α )} \Rightarrow 2 \int \frac{d t}{( β - α ) - t ^{2}}$

$\Rightarrow 2 sin^{- 1} (\frac{t}{β - α}) \Rightarrow [2 sin^{- 1} (\frac{x - α}{β - α})]_{α}^{β}$

$\Rightarrow 2 (\frac{π}{2} - 0) \Rightarrow π$

Example 3: $\int_{- π /2}^{π /2} sin^{4} x cos^{6} x d x =$

(A) $\frac{3 π}{64}$

(B) $\frac{3 π}{572}$

(C) $\frac{3 π}{256}$

(D) $\frac{3 π}{128}$

Ans. (C)

Solution:

$I = \int_{- π /2}^{π /2} sin^{4} x cos^{6} x d x = 2 \int_{0}^{π /2} sin^{4} x cos^{6} x \cdot d x = 2 \frac{( 3.1 ) ( 5 \cdot 3 \cdot 1 )}{10.8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{π}{2} = \frac{3 π}{256}$

Example 4: $\int_{0}^{1} \frac{t a n ^{- 1} x}{x} d x = λ \int_{0}^{π /2} \frac{θ}{s i n θ} d θ$ , then find λ.

Ans. 1/2

Solution:

$L.H.S. \int \frac{t a n ^{- 1} x}{x} d x$

Put

$= \int_{0}^{π /4} \frac{t}{t a n t} \cdot sec^{2} t d t = \int_{0}^{π /4} \frac{t}{s i n t c o s t} d t = \int_{0}^{π /4} \frac{2 t}{s i n 2 t} d t$

2t = θ

2dt= dθ

$= \frac{1}{2} \int_{0}^{π /2} \frac{θ}{s i n θ} d θ$

∴ $λ = \frac{1}{2}$

Example 5: Evaluate $\int_{2}^{8} ∣ x - 5∣ d x$ .

Solution:

$\int_{2}^{8} ∣ x - 5∣ d x = \int_{2}^{5} (- x + 5) d x + \int_{5}^{8} (x - 5) d x = 9$

Example 6: $Evaluate \int_{- 1}^{1} \frac{3 ^{x} + 3 ^{- x}}{1 + 3 ^{x}} d x$

Solution:

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

$= \int_{0}^{1} (3^{x} + 3^{- x}) d x = (\frac{3 ^{x}}{l n 3} - \frac{3 ^{- x}}{l n 3})_{0}^{1} = (\frac{3}{l n 3} - \frac{3 ^{- 1}}{l n 3}) - (\frac{1}{l n 3} - \frac{1}{l n 3}) = \frac{1}{l n 3} [3 - \frac{1}{3}] = \frac{8}{3 l n 3}$

Example 7: $Prove that \int_{0}^{π /2} lo g (sin x) d x = \int_{0}^{π /2} lo g (cos x) d x = - \frac{π}{2} lo g 2$

Solution:

Let $I = \int_{0}^{π /2} lo g (sin x) d x$ ...(i)

then $I = \int_{0}^{π /2} lo g sin (\frac{π}{2} - x) d x = \int_{0}^{π /2} lo g (cos x) d x$ ...(ii)

adding (i) and (ii), we get

$2 I = \int_{0}^{π /2} lo g sin x d x + \int_{0}^{π /2} lo g cos x d x = \int_{0}^{π /2} (lo g sin x + lo g cos x) d x$

$\Rightarrow 2 I = \int_{0}^{π /2} lo g (sin x cos x) d x = \int_{0}^{π /2} lo g (\frac{2 s i n x c o s x}{2}) d x$

$= \int_{0}^{π /2} lo g (\frac{s i n 2 x}{2}) d x = \int_{0}^{π /2} lo g (sin 2 x) d x - \int_{0}^{π /2} (lo g 2) d x = \int_{0}^{π /2} lo g sin 2 x \cdot d x - (lo g 2) (x)_{0}^{π /2}$

$\Rightarrow 2 I = \int_{0}^{π /2} lo g (sin 2 x) d x - \frac{π}{2} lo g 2$ ...(iii)

Let

$I_{1} = \int_{0}^{π /2} lo g (sin 2 x) d x$ ,

putting 2 x=t, we get

$I_{1} = \int_{0}^{π} lo g (sin t) \frac{d t}{2} = \frac{1}{2} \int_{0}^{π} lo g (sin t) d t = \frac{1}{2} \cdot 2 \int_{0}^{π /2} lo g (sin t) d t$

$I_{1} = \int_{0}^{π /2} lo g (sin x) d x$

∴ (iii) becomes; $2 I = I - \frac{π}{2} lo g 2$

Hence $\int_{0}^{π /2} lo g sin x d x = - \frac{π}{2} lo g 2$

Example 8: Find the slope of the tangent to the curve $y = \int_{x}^{x^{2}} cos^{- 1} t^{2} d t at x = \frac{1}{4 2}$

Solution:

Given curve is

$y = \int_{x}^{x^{2}} cos^{- 1} t^{2} d t; \frac{d y}{d x} = \frac{d}{d x} \int_{x}^{x^{2}} cos^{- 1} t^{2} d t$

using Leibnitz theorem, $\frac{d y}{d x} = 2 x cos^{- 1} x^{4} - cos^{- 1} x^{2}$

$(\frac{d y}{d x})_{x = \frac{1}{4 2}} = \frac{2}{2 ^{1/4}} cos^{- 1} \frac{1}{2} - cos^{- 1} \frac{1}{2} = 2^{3/4} \frac{π}{3} - \frac{π}{4} = (\frac{4 8}{3} - \frac{1}{4}) π$

9.0Practice Problems on Definite Integration

a. If $f (x) = ∣ x ∣ + ∣ x - 1∣ + ∣ x - 2∣, x \in R then \int_{0}^{3} f (x) dx =$

b. Evaluate: $\int_{0}^{1} \frac{x ^{2} - 1}{1 + x ^{2} + x ^{4}} d x$

c. Evaluate: $\int_{0}^{π /2} \frac{d x}{1 + c o s θ \cdot c o s x} θ \in (0, π)$

d. The value of the integral $\int_{- 1}^{1} lo g (x + x^{2} + 1) d x$ is:

e. Evaluate: $\int_{- 1}^{1} \frac{( 2 x ^{332} + x ^{998} + 4 x ^{1668} \cdot s i n x ^{691} )}{1 + x ^{666}} d x$

Answers:

a. $\frac{19}{2}$

b. $\frac{1}{2} ln (\frac{1}{3})$

c. $\frac{θ}{s i n θ}$

d. 0

e. $\frac{2}{333} [\frac{π}{4} + 1] = \frac{π + 4}{666}$

10.0Solved Questions on Definite Integration

1. How do you define a definite integral?

Ans: A definite integral is defined as the integral of a function f(x) from a to b, denoted as $\int_{a}^{b} f (x) d x$ , where 'a' and 'b' are the limits of integration.

2. How do you perform definite integration by parts?

Ans: Integration by parts for definite integrals is given by: $\int_{a}^{b} u d v = uv ∣_{a}^{b} - \int_{a}^{b} v d u$ where u and v are differentiable functions.

3. How do you evaluate a definite integral using the limit of a sum?

Ans: A definite integral can be evaluated using Riemann sums, which approximate the area under the curve by summing the areas of rectangles:

$\int_{a}^{b} f (x) d x = lim_{n \to \infty} \sum_{i = 1}^{n} f (x_{i}^{*}) Δ x$ , where $Δ x = \frac{b - a}{n} and x_{i}^{*}$ is a sample point in the i^th subinterval.

Frequently Asked Questions

What is Definite Integration?

What are the common rules for definite integrals?

Frequently Asked Questions

What is Definite Integration?

What are the common rules for definite integrals?

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Related Articles:-

Indefinite Integration

Double Integral

Integrals of Particular Functions

Important Integration Formulas

Complex Numbers

Chain Rule Questions

Coplanar Vectors

Definite Integration

1.0Definite Integration Definition

2.0Fundamental Theorem of Calculus

3.0Properties of Definite Integrals

4.0Walli’s Theorem

5.0Derivative of Antiderivatives(Newton-Leibnitz Theorem)

6.0Definite Integration as the limit of Sum

7.0Estimation of Definite Integral and General Inequality

8.0Definite Integration Questions

9.0Practice Problems on Definite Integration

10.0Solved Questions on Definite Integration

Table of Contents

Join ALLEN!

Related Articles:-

Indefinite Integration

Double Integral

Integrals of Particular Functions

Important Integration Formulas

Complex Numbers

Chain Rule Questions

Coplanar Vectors

Definite Integration

1.0Definite Integration Definition

2.0Fundamental Theorem of Calculus

3.0Properties of Definite Integrals

4.0Walli’s Theorem

5.0Derivative of Antiderivatives(Newton-Leibnitz Theorem)

6.0Definite Integration as the limit of Sum

7.0Estimation of Definite Integral and General Inequality

8.0Definite Integration Questions

9.0Practice Problems on Definite Integration

10.0Solved Questions on Definite Integration

Table of Contents