Calculus

Integration and Differentiation

Calculus involves two of the most important concepts - integration and differentiation. Integration is used to calculate the area of a plane figure, while differentiation is a process of finding a function that outputs the rate of change of one variable with respect to another. Integration is the reverse of differentiation, and is also referred to as the antiderivative.

In this section, aspirants will learn about indefinite and definite integration, a list of important formulas, how to use integral properties to solve integration problems, and various integration methods.

Indefinite Integration

The indefinite integral is defined as a function that will calculate the area under the curve of a given function from an unspecified point to a specified point. Read more.

Standard Formulas for Indefinite Integration

1. (\int{{{x}^{n}}dx=\frac{{{x}^{n+1}}}{n+1}+c})

2. $\int a^x , dx = \frac{a^x}{\ln a} + c$

3. $$\int{{e}^{x}}dx = {e}^{x} + c$$

4. $$\int{\frac{1}{x}}dx=\ln\,x+c$$

5. $$\int{\sin x\,dx=-\cos x+c}$$

6. (\int{\cos x,dx=\sin x+C})

7. $\int{{{\sec }^{2}}x,,dx=\tan x+C}$

8. (\int{\cos e^{{c}^{2}}x,,dx=-\cot x,+c})

9. (\int{\sec x\tan x,dx=\sec x - \ln|\sec x + \tan x| + c})

10. $$\int{\cos ec\,\,x\,\,}\cot x\,dx=-\cos ec\,x+c$$

$$\int{\frac{1}{\sqrt{1-{{x}^{2}}}}dx = \left{ \begin{matrix} {{\sin }^{-1}}x+c \ -{{\cos }^{-1}}x+c \ \end{matrix} \right.$$

12. $$\int{\frac{1}{1+{{x}^{2}}}}dx = \left{ \begin{matrix} {{\tan }^{-1}}x+c \ -{{\cot }^{-1}}x+c \ \end{matrix} \right.$$

13. $$\frac{1}{x\sqrt{{{x}^{2}}-1}} = \left{\begin{matrix} \sec^{-1}x + c \ -\csc^{-1}x + c \end{matrix} \right.$$

⇒ Introduction to Integration: Learn More

Methods of Integration

1. Substitution Method

2. Integration by Partial Fractions

1. By Parts
2. By Parts
3. By Parts

4. Euler Substitution

5. Reduction Method

Benefits of Indefinite Integration

1. (\begin{array}{l}k\int{f\left( x \right)dx = k\int{f\left( x \right)dx}}\end{array} )

2. (\int{f\left( x \right)+g\left( x \right) , dx = \int{f\left( x \right) , dx + \int{g\left( x \right) , dx}})

Illustrations of Integration Properties and Substitution Method:

Illustration 1: Solve $$\int{\frac{{{\left( 1+x \right)}^{3}}}{\sqrt[3]{{{x}^{2}}}}}dx$$

Solution: (\int\frac{(1+x)^3}{x^{2/3}}dx=\int\frac{1+3x+3x^2+x^3}{x^{2/3}}dx)

\(\int({{{x}^{5/3}}+3{{x}^{2/3}}+3{{x}^{7/3}}}+{{x}^{10/3}})dx\)

(\begin{array}{l}3{{x}^{1/3}}+\frac{9}{4}{{x}^{4/3}}+\frac{9}{7}{{x}^{7/3}}+\frac{3}{10}{{x}^{10/3}}+c\end{array})

Illustration 2: Solve (\int{{\sec }^{2}}x\cos e{{c}^{2}}x,,dx)

Solution: (\int{{{\sec }^{2}}x\cos e{{c}^{2}}x,,dx=\int{\frac{1}{{{\sin }^{2}}x,{{\cos }^{2}}x}}dx})

(\int{{{\sec }^{2}}x+\cos e{{c}^{2}}x,,dx})

\(\tan x - \cot x + c\)

Illustration 3. $$\int{2x,,\sin \left( {{x}^{2}} \right)dx}$$

Solution: x² = t.

2x \frac{dx}{dt} = 1

(\int{\sin t,,dt=-\cos t+C})

\(-\cos^2 x + c\)

Illustration 4. (\int{{{\sin }^{3}}x{{\cos }^{5}}x,,dx})

Solution: (\int {{{\sin }^2}x{{\cos }^5}x\sin x,dx} )

‘(\int{\left( 1-{{\cos }^{2}}x \right)}{{\cos }^{5}}x,,\sin x,,dx)’

Put cos x = t

Therefore, $\int -\sin x ,dx = t + C$

So, new equation is: $$\int{\left( 1-{{t}^{2}} \right)}{{t}^{5}}\left( -dt \right)$$

‘(\int{{{t}^{5}}-{{t}^{7}}}dt = \frac{1}{6}{t}^{6} - \frac{1}{8}{t}^{8} + C)’

(-\left( \frac{{{t}^{6}}}{6}-\frac{{{t}^{8}}}{8} \right))

Re-substituting the value, we get:

(\frac{{{\cos }^{6}}x}{6} + \frac{{{\cos }^{8}}x}{8} + c)

Important Formulae Set for Indefinite Integration

1. (\int{\frac{dx}{\sqrt{a^2 - x^2}}} = \sin^{-1}\left(\frac{x}{a}\right) + c)

2. $$\int\frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right|$$

3. $$\int{\frac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}} = \ln\left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|$$

4. $$\int{\frac{1}{{{a}^{2}}+{{x}^{2}}}}dx=\frac{1}{a}{{\tan }^{-1}}\left( \frac{x}{a} \right).$$

5. $$\int{\frac{1}{{a^2}-{x^2}}}dx = \frac{1}{2a}\ln\left| \frac{a+x}{a-x} \right|.$$

6. $$\int{\frac{1}{{{x}^{2}}-{{a}^{2}}}}dx=\frac{1}{2a}\ln\left| \frac{x-a}{x+a} \right|.$$

7. $$\int{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx = \frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \frac{x}{a} \right).$$

$$\int{\sqrt{{{a}^{2}}+{{x}^{2}}}},dx = \frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}} + \frac{{{a}^{2}}}{2}\ln\left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|$$

$$\int{\sqrt{{{x}^{2}}-{{a}^{2}}}}dx = \frac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}} + \frac{{{a}^{2}}}{2}\ln\left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|$$

Types of Substitutions:

Type I:

Type I

(\int{{{\sin }^{m}}x{{\cos }^{n}}x,,dx})

Rule: All employees must wear a uniform while on duty.

1. If m and n are both odd, then substitute for an even power.

If both numbers are odd, substitute either one.

3. If both are even, use trigonometric identities.

Type II: Substitution of Trigonometric Functions

1. (\begin{array}{l}x=a\sin \theta\rightarrow{{}}\sqrt{{{a}^{2}}-{{x}^{2}}}\end{array} )

2. (\begin{array}{l}x=a\tan \theta \\ \sqrt{{{x}^{2}}+{{a}^{2}}}=a\end{array})

3. (\begin{array}{l}{x} = a \sec \theta \\sqrt{{{x}^{2}} - {{a}^{2}}} \rightarrow{} \end{array})

4. (\begin{array}{l}x=a{{\cos }^{2}}\theta +b{{\sin }^{2}}\theta \rightarrow{{}}\sqrt{\frac{x-a}{b-x}}\end{array})

5. (\displaystyle \sqrt{\frac{x-a}{x-b}} \rightarrow x = a\sec^2\theta - b)

Type III:

(\begin{array}{l} \int{\frac{px+q}{a{{x}^{2}}+bx+c}},dx \\ \int{\frac{px+q}{\sqrt{a{{x}^{2}}+bx+c}}},dx \\ \int{\left( px+q \right)}\sqrt{a{{x}^{2}}+bx+c},dx \end{array})

We express (px + q) as: (\begin{array}{l}m{{\left( a{{x}^{2}}+bx+c \right)}^{1}}+n\end{array} ) Then, this gets changed to the standard integral form (\int m{{\left( a{{x}^{2}}+bx+c \right)}^{1}}+n,dx).

Biquadratic Substitutions

1. (\int{f\left( x+\frac{1}{x} \right)}\left( 1-\frac{1}{{{x}^{2}}} \right)dx )

(\begin{array}{l}\text{Solve for } x: \ x+\frac{1}{x}=t \end{array})

2. (\int f\left( x \right) \left(1+\frac{1}{x^2}\right) - \int\frac{f\left(x\right)}{x^2} dx)

(\begin{array}{l}\text{Solve}\ \left( x-\frac{1}{x} \right)=t\end{array} )

Solved Problems:

Problem 1: Solve (\int{\frac{dx}{\sqrt{{{x}^{2}}+2x+2}}})\

Solution: (\int{\frac{dx}{\sqrt{{{\left( x+1 \right)}^{2}}+1}}} = \int{\frac{dx}{\sqrt{{{x}^{2}}+2x+2}}}).

$\log \left| \left( x+1 \right) + \sqrt{{{\left( x+1 \right)}^{2}}+1} \right|$

Problem 2: Solve (\int{\frac{5{{x}^{2}}+2x+3}{\sqrt{5{{x}^{2}}+2x+3}},,dx)

Solution: (\int{\frac{5}{2}{{x}^{2}}+x+3},,dx)

Here

(\begin{array}{l}5{{\left( x+\frac{1}{5} \right)}^{2}}+{{\left( \sqrt{\frac{14}{25}} \right)}^{2}} = 5{{x}^{2}}+2x+3\end{array})

(\int{\sqrt{5{{x}^{2}}+2x+3},dx}=\sqrt{5}\int{{{\left( x+\frac{1}{5} \right)}^{2}}+{{\left( \sqrt{\frac{14}{25}} \right)}^{2}},dx})

\(\sqrt{5}\frac{\left( x+\frac{1}{5} \right)}{2}\sqrt{\left( x+\frac{1}{5} \right)^{2}+\left( \sqrt{\frac{14}{15}} \right)^{2}}+\frac{14}{25.2}\log \left| \left( x+\frac{1}{5} \right)+\sqrt{\left( x+\frac{1}{5} \right)^{2}+\frac{14}{15}} \right|\)

Problem 3: Evaluate (\int{\frac{x+1}{{{x}^{2}}+3x+4}}dx)

Solution: $$\int{\frac{x+1}{{{x}^{2}}+3x+4}}dx$$

(\begin{array}{l}x+1 = a{{\left( {{x}^{2}} + 3x + 4 \right)}^{1}} + b\end{array})

(\begin{array}{l}2ax + 3a + b\end{array})

(\therefore a=\frac{1}{2},,,,,b=\frac{-1}{2})

(\begin{array}{l}\Rightarrow I=\int \frac{1}{2}\left(1 - \frac{1}{{{x}^{2}}+3x+4} \right)\ d\left( {{x}^{2}}+3x+4 \right)\end{array} )

(\frac{1}{2}\log \left( {{x}^{2}}+3x+4 \right) - \frac{1}{2} \left[ \frac{2}{3\left( x+\frac{3}{2} \right)} + \frac{7}{4}x + c \right])

(\frac{1}{2}\log \left( {{x}^{2}}+3x+4 \right)-\frac{1}{2}\sqrt{7}\tan ^{-1}\left( \frac{x+\frac{3}{2}}{\sqrt{7}} \right))

(\frac{1}{2}\log \left| {{x}^{2}}+3x+4 \right|-\frac{1}{\sqrt{7}}{{\tan }^{-1}}\left( \frac{2x+3}{\sqrt{7}} \right))

Problem 4: Solve [ \int{\frac{1+{{x}^{2}}}{1+{{x}^{4}}}},dx ]

Solution: (\int{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)}{\left( 1+\frac{1}{{{x}^{2}}}+{{x}^{2}} \right)}},dx)

(\int{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)dx}{{{\left( x-\frac{1}{x} \right)}^{2}}+2}})

(\int{\frac{d\left( x-\frac{1}{x} \right)}{{{\left( x-\frac{1}{x} \right)}^{2}}+2}} = \frac{1}{2}\ln{\left| x-\frac{1}{x} \right|} + C)

(\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{\sqrt{2}\left( x-\frac{1}{x} \right)}{2} \right))

Integration by Partial Fractions

Integrals of rational functions can be evaluated by splitting them into partial fractions.

(\frac{f\left( x \right)}{g\left( x \right)})

A function of the form $$\frac{f(x)}{g(x)}$$ where $f$ and $g$ are polynomials is called a rational function.

If the degree of f is less than the degree of g, it is referred to as a proper fraction.

Otherwise, it is an improper fraction.

Then $$\frac{f\left( x \right)}{g\left( x \right)} = h\left( x \right) + \frac{d\left( x \right)}{g\left( x \right)},$$ where the degree of $d$ is less than the degree of $g$.

Cases:

1. When $g(x)$ is expressed as the product of non-repeating linear factors.

(\begin{array}{l}g\left( x \right)=\prod\limits_{i=1}^{n}{\left( x-{{a}_{i}} \right)} \end{array} )

(\begin{array}{l}\frac{f}{g}=\frac{{A}_1}{x-{a}_1}+\frac{{A}_2}{x-{a}_2}+\dots + \frac{{A}_n}{x-{a}_n}\end{array})

2. Some factors of $g$ are repeating, then $$g(x) = (x-a)^k(x-a_1)(x-a_2)…$$

(\frac{f}{g}=\frac{{A_1}}{\left(x-a\right)}+\frac{{A_2}}{{\left(x-a\right)}^2}+\dots+\frac{{A_k}}{{\left(x-a\right)}^n}+\dots )

3. (\frac{f}{g}=\frac{Ax+B}{a{{x}^{2}}+bx+c}) if g(x) has a quadratic term

Where A and B are constants determined by comparing coefficients.

Let’s Explore Examples

Example 1: Solve (\int{\frac{dx}{\left( x+1 \right)\left( x-2 \right)}})

Answer: (\int{\frac{dx}{\left( x+1 \right)\left( x-2 \right)}} = \frac{1}{2} \ln\left| \frac{x+1}{x-2} \right| + C)

Solution: (\int{\frac{a\left( x-2 \right)+b\left( x+1 \right)}{\left( x+1 \right)\left( x-2 \right)}},dx = a\ln{\left| x+1 \right|}+b\ln{\left| x-2 \right|} + c)

\(\int{\frac{-\frac{1}{3}}{x+1}}\,dx + \int{\frac{\frac{1}{2}}{x-2}}\,dx\)

(\begin{array}{l}=-\frac{1}{3}\ln\left| x+1 \right| + \frac{1}{2}\ln\left| x-2 \right|\end{array})

Example 2: Solve \begin{align*}\int{\frac{2{{x}^{2}}-1}{\left( x-1 \right){{\left( x+1 \right)}^{2}}}}dx\end{align*}

Solution: (\int{\frac{A\left( x+1 \right)+B\left( x-1 \right)}{\left( x-1 \right)\left( x+1 \right)^{2}}+C\frac{1}{\left( x+1 \right)^{2}}}dx=2{x}^{2}-1 )

(\begin{array}{l}A{{\left( x+1 \right)}^{2}}+B\left( x+1 \right)\left( x+1 \right)+C\left( x-1 \right)=2{{x}^{2}}-1\end{array})

x = 1 \Rightarrow A = \frac{1}{2}

x = -1 \Rightarrow C = -\frac{1}{2}

B = \frac{3}{2}.

(\begin{array}{l}\int{\frac{1}{2}\frac{1}{x-1}}dx+\frac{3}{2}\int{\frac{1}{x+1}}dx+\left( \frac{-1}{2} \right)\int{\frac{1}{{{\left( x-2 \right)}^{2}}}}dx\end{array} )

(\frac{1}{2}\ell n\left| x-1 \right| + \frac{3}{2}\ell n\left| x+1 \right| + \frac{1}{2}\frac{1}{x+2} + c )

Example 3: Solve $$\int{\frac{dx}{\left( x+2 \right)\left( {{x}^{2}}+1 \right)}}$$

Solution: (\int{\frac{A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)\left( x+2 \right)}{\left( x+2 \right)\left( {{x}^{2}}+1 \right)}}dx=A+\frac{Bx+C}{{{x}^{2}}+1})

x = -2 ⇒ A = 1/5

B = -1/2 * x^2 and C = 2/5 * x^2

On substituting the values and integrating, we have.

(\begin{array}{l}I=\frac{1}{5}\ln\left| x+2 \right|-\frac{1}{10}\ln\left( {{x}^{2}}+1 \right)+\frac{2}{5}\arctan\left( x \right)\end{array} )

Integration of Trigonometric Functions

Type 1: (\int {{{\sin }^{m}}x{{\cos }^{n}}dx} )

1. If m is odd, put cos x = t

2. If n is odd, let sin x = t

If m and n are rational numbers, then put $\tan x = t$

4. If both are even, then use the reduction method.

For instance:

(\int{\frac{{{\cos }^{3}}x}{{{\sin }^{6}}x}dx=\int{\frac{1-{{t}^{2}}}{{{t}^{6}}}dt}})

t = sin(x)

(\int{{{t}^{-6}}-{{t}^{-4}}dt} = \frac{{{t}^{-5}}}{5} - \frac{{{t}^{-3}}}{3} + C)

(\begin{array}{l}=\frac{-1}{5\sin^5x} + \frac{1}{3\sin^3x} + c\end{array})

Type 2: (\int{\frac{dx}{a\cos x + b\sin x + c}})

t = \tan \left(\frac{x}{2}\right)

Illustration:

(\begin{array}{l}\int{\frac{dx}{2+\sin x}} = \int{\frac{dt}{1+t^2}}\end{array} ) (\begin{array}{l}\Rightarrow t=\tan \left( \frac{x}{2} \right)\end{array} )

(\frac{d}{dt}\left( {{t}^{2}} \right)=2t)

(\begin{array}{l}=\int{\frac{d(1+{{t}^{2}})}{2(1+{{t}^{2}})+2t}}\end{array} ) (\begin{array}{l}\Rightarrow \int{\frac{d(1+{{t}^{2}})}{{{t}^{2}}+t+1}}\end{array} )

(\frac{2}{\sqrt{3}}\arctan\left(\frac{2t+1}{\sqrt{3}}\right))

(\frac{2}{\sqrt{3}}\arctan\left(\frac{2\tan\left(\frac{x}{2}\right)+1}{\sqrt{3}}\right) + c)

Type 3

\(\int{\frac{dx}{a{{\cos }^{2}}x+b{{\sin }^{2}}x}}\), \(\int{\frac{dx}{a+b{{\sin }^{2}}x}}\)

(\begin{array}{l}\int{\frac{1}{a+b{{\cos }^{2}}x}},dx, \\int{\frac{1}{{{\left( a\sin x+b\cos x \right)}^{2}}}},dx \\text{or} \\int{\frac{1}{a+b{{\sin }^{2}}x+{{\cos }^{2}}x}},dx\end{array} )

Rule: All student must have their ID cards with them at all times.

Divide both the numerator and denominator by $\cos^2x$.

Illustration:

(\begin{array}{l}\int{\frac{dx}{2+\sin x}} = \int{\frac{d\left(\tan\left(\frac{x}{2}\right)\right)}{2+\tan\left(\frac{x}{2}\right)}}\end{array} ) (\begin{array}{l}\Rightarrow \int{\frac{dx}{2+\sin x}} = \int{\frac{dt}{2+t}}\end{array} )

\(\frac{d}{dt}\left(t^2 + 1\right) = 2dt\)

(\begin{array}{l}=\int{\frac{d(2t)}{(1+{{t}^{2}})(2+\frac{2t}{1+{{t}^{2}}})}}\end{array} ) (\begin{array}{l}\Rightarrow \int{\frac{2dt}{(1+{{t}^{2}})(2+\frac{2t}{1+{{t}^{2}}})}}\end{array} ) (\begin{array}{l}\Rightarrow \int{\frac{dt}{{{t}^{2}}+t+1}}\end{array} )

(\frac{2}{\sqrt{3}}\arctan\left(\frac{2t+1}{\sqrt{3}}\right))

(\frac{2}{\sqrt{3}}\arctan\left(\frac{2\tan\frac{x}{2} + 1}{\sqrt{3}}\right) + c)

Type 3:

‘\(\int{\frac{dx}{a{{\cos }^{2}}x+b{{\sin }^{2}}x}},,\int{\frac{dx}{a+b{{\sin }^{2}}x}}\)’

(\begin{array}{l}\int{\frac{1}{a+b{{\cos }^{2}}x}},dx,\int{\frac{1}{a+b{{\sin }^{2}}x+{{\cos }^{2}}x}},dx\end{array} ) or (\begin{array}{l}\int{\frac{1}{{{\left( a\sin x+b\cos x \right)}^{2}}}},dx\end{array} )

Rule: All employees must wear a name tag while on the job.

Divide both the numerator and denominator by $\cos^2 x$.

Illustration:

(\int{\frac{1}{{{\cos }^{2}}x}},dx=\int{\frac{\frac{1}{{{\cos }^{2}}x}}{\frac{3}{{{\cos }^{2}}x}-\frac{4{{\sin }^{2}}x}{{{\cos }^{2}}x}}},dx)

\(\int{\frac{1}{3-4\tan^2 x}\sec^2 x\,dx}\)

(\begin{array}{l}=\int{\frac{sec^2x \ dx}{3\left( 1+{{tan^2x}} \right)-4{{tan^2x}}}}\end{array} )

‘(\displaystyle \int{\frac{dt}{3-{{t}^{2}}}} )’

\(\int{\frac{1}{{{\left( \sqrt{3} \right)}^{2}}-{{t}^{2}}}}dt\)

(\frac{1}{2\sqrt{3}} \ln \left| \frac{\sqrt{3} + t}{\sqrt{3} - t} \right|)

(\begin{array}{l} \log \left| \frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x} \right| = 2\sqrt{3}\cdot \frac{1}{2\sqrt{3}}\end{array})

Indefinite Integration - Video Lesson

![Indefinite Integration - Video Lesson]()

Indefinite Integration Problems

Definite Integrals

Divide [a, b] into $n$ parts of width $h = \frac{b - a}{n}$, creating $(n - 1)$ partitions of the interval.

(\int\limits_{a}^{b}f(x) , dx = \underset{h \to 0}{\mathop{\lim }},h,\sum\limits_{n=0}^{n-1}f(a + rh))

(\int\limits_{a}^{b}{f\left( x \right),dx} = \underset{h \to 0}{\mathop{\lim }},h,\sum\limits_{n = 0}^{n - 1}{f\left( a + rh \right)}).

Definition of the Definite Integral as the Limit of a Sum

(\lim_{h\to 0} h \left[f\left(a\right) + f\left(a+h\right) + f\left(a+2h\right) + \dots + f\left(a + \left(n-1\right)h\right)\right])

‘(\begin{array}{l}nh = b - a\end{array})’

Integrals using Limits

Limit of a Sum Evaluation

(\begin{array}{l}f\left(x\right)=x \\ h=\frac{b-a}{n} \\ I=\int\limits_{a}^{b}f\left(x\right)dx\end{array})

f(a) = a

f(a + h) = a + h

(\begin{array}{l}I=h\left[ a+\left( a+h \right)+….\left( a+\left( n-1 \right)h \right) \right] = nh\end{array})

\(\displaystyle h\sum_{i=1}^{n-1} ih \)

(\begin{array}{l} hna + \frac{h^2 n(n - 1)}{2} \end{array})

(\begin{array}{l}\frac{{{b}^{2}}-{{a}^{2}}}{2}=\left( b-a \right)a+\frac{{{h}^{2}}{{n}^{2}}}{2}-\frac{{{\left( b-a \right)}^{2}}}{2}\end{array})

(\sum\limits_{r=1}^{n}{n} = \frac{n(n+1)}{2})

(\sum_{r=1}^{n}{n^2} = \frac{n(n+1)(2n+1)}{6})

(\sum\limits_{r=1}^{n}{n^{3}} = \left[ \frac{n\left( n+1 \right)}{2} \right]^{2})

GP … $$\begin{array}{l}a+ar+a{{r}^{2}}….a{{r}^{n-1}}=\frac{a\left( 1-{{r}^{n}} \right)}{1-r}\end{array} $$

(\sin \alpha + \sin \left(\alpha + \beta \right) + \sin \left(\alpha + 2\beta \right) + \dots + \sin \left(\alpha + \left(n - 1\right)\beta \right))

(\frac{\sin \left( n\frac{\beta }{2} \right)}{\sin \left( \frac{\beta }{2} \right)} \sin \left[ \frac{\alpha + (n-1)\beta}{2} \right])

\(\frac{\sin \left( n\frac{diff}{2} \right)}{\sin \left( \frac{diff}{2} \right)} \cdot \sin \left( \frac{1st+last}{2} \right)\)

(\cos \left( \frac{1st+last}{2} \right) \cdot \frac{\sin \left( \frac{n,diff}{2} \right)}{\sin \left( \frac{diff}{2} \right)})

(\begin{array}{l}\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}-\frac{1}{{{6}^{2}}}-\frac{1}{{{1}^{2}}}=\frac{{{\pi }^{2}}}{12}\end{array})

(\begin{array}{l}\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}+\cdots =\frac{{{\pi }^{2}}}{12}\end{array})

(\int\limits_{a}^{b}{e^{x},dx=\underset{n\to \infty }{\mathop{\lim }},h\left[e^{a}+e^{a+h}+e^{a+2h}+\cdots+e^{a+(n-1)h}\right])

(\begin{array}{l}h^a \times \left[ 1 + e^n + e^{2h} + \cdots \right]\end{array})

(\frac{e^a \left[ e^{nh}-1 \right] h}{e^n - 1})

(\begin{array}{l}{{e}^{b}}-{{e}^{a}}\end{array})

(\int\limits_{a}^{b} \sin x,dx=\underset{n\to \infty}{\mathop{\lim }},h\left[ \sin a + \sin\left(a + n\right) + \cdots + \sin\left(a + \left(n - 1\right)h\right) \right])

(\underset{n\to \infty }{\mathop{\lim }},h\frac{\sin \left( \frac{nh}{2} \right)}{\sin \left( \frac{h}{2} \right)}\left[ \sin \left( \frac{2a+(n-1)h}{2} \right) \right])

$$\begin{array}{l}=\frac{\sin \left( \frac{b-a}{2} \right)}{\left( \frac{1}{h} \right)\sin \left( \frac{h}{2} \right)}\left[ \sin \left( a+\frac{nh}{2}-\frac{h}{2} \right) \right]\=2\sin \left( \frac{b-a}{2} \right)\sin \left( \frac{a+b}{2} \right)\end{array}$$

(\cos \left( \frac{b-a}{2} \right)-\cos \left( \frac{a+b}{2} \right))

cos a - cos b

(\begin{array}{l}\int\limits_{0}^{2}{\left( 3{{x}^{2}}+2x+1 \right)dx=\frac{14}{3}}\end{array} )

(\begin{array}{l}\int\limits_{0}^{\frac{\pi }{2}}{\cos x,dx=1} \\int\limits_{0}^{\frac{\pi }{2}}{{{\sin }^{2}}x,dx=h\left[ \frac{1-\cos 2x}{2}+\frac{1-{{\cos }^{\left( 2n+2n \right)}}}{2}…… \right]}\end{array} )

(\frac{1}{2}\left( b-a \right) - \frac{1}{4}\left( \sin 2b - \sin 2a \right))

Definite Integration and Its Limits

(\int\limits_{a}^{b}{f\left(x\right) , dx = \underset{h \to 0}{\mathop{\lim }},\sum\limits_{r=1}^{n}{f\left(a + rh\right)}})

Special case: a = 0 and b = 1

(\int\limits_{0}^{1}{f\left( x \right),dx=\underset{h\to 0}{\mathop{\lim }},\frac{1}{n}\sum\limits_{r=1}^{n}{f\left( \frac{r}{n} \right)}).

(\int\limits_{0}^{k}{f\left(x\right),dx=\underset{h\to 0}{\mathop{\lim }},\frac{1}{n}\sum\limits_{r=1}^{kn}{f\left(\frac{r}{n}\right)}).

(\begin{array}{l}\sum\rightarrow \text{Multiple of }n\ \int \frac{1}{n}, dx \Rightarrow \frac{r}{n}\end{array})

Using Limits to Solve Integrals:

1. Identify the function to be integrated
2. Identify the limits of integration
3. Rewrite the integral as a limit
4. Evaluate the limit
5. Interpret the answer

(\underset{n\to \infty }{\mathop{\lim }},\left[ \frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}+\cdots+\frac{1}{nb} \right])

(\underset{n\to \infty }{\mathop{\lim }},\sum\limits_{r=0}^{b-a}{\frac{1}{n(a+r)}})

(\begin{array}{l}=\sum\limits_{n\to \infty }{\frac{1}{n}\sum\limits_{r=0}^{\left( b-a \right)n}{\frac{1}{a+\frac{r}{n}}}}\ =\int\limits_{0}^{b-a}{\frac{dx}{a+x}}\ =\left[ \log \left( a+x \right) \right]_{a}^{b-a}\ =\log \left( \frac{b}{a} \right)\end{array} )

Videos on Applications of Integrals

Definite Integral and Area Under the Curve: Important Topics

Important Questions on Definite Integral and Area Under the Curve

![Definite Integral & Area Under the Curve - Important Questions]()

#Most Important Questions from Definite Integration for JEE Advanced

Frequently Asked Questions

Integration refers to the combining of different components or subsystems into one system to create a unified whole.

Finding the antiderivative of a function is a method of integration.

What are the Rules of Integration?

-There are many rules of integration, such as:

• The Power Rule
• The Sum and Difference Rules
• The Exponential Rule
• The Reciprocal Rule
• The Constant Rule
• The Substitution Rule
• The Rule of Integration by Parts, etc.

What are the Benefits of Integration?

Integral calculus originated from the concept of integration in mathematics, which is used to find areas, volumes, displacement, etc.