Integration by Parts MCQ Quiz in मल्याळम - Objective Question with Answer for Integration by Parts - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Calculus:

Derivatives of Trigonometric Functions:

\(\rm \frac{d}{dx}\sin x=\cos x\ \ \ \ \ \frac{d}{dx}\cos x=-\sin x\) 

\(\rm \frac{d}{dx}\tan x=\sec^2x\ \ \ \ \ \ \frac{d}{dx}\cot x=-\csc^2 x\)

\(\rm \frac{d}{dx}\sec x=\tan x\sec x\ \ \ \frac{d}{dx}\csc x=-\cot x\csc x\)

Integration by Parts:

∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx.

Calculation:

Let I = \(\rm\int x\sin x\ dx\).

Considering x as the first function and sin x as the second function, we get:

⇒ I = x ∫ sin x dx - ∫ [(1) ∫ sin dx] dx

⇒ I = x (-cos x) - ∫ (-cos x) dx

⇒ I = sin x - x cos x + C.

Concept:

Integration by parts: Integration by parts is a method to find integrals of products.

The formula for integrating by parts is given by:

⇒ \(\rm ∫ u vdx=u ∫ vdx- ∫ \left({du\over dx}\times ∫ vdx\right)dx \) + C

where u is the function u(x) and v is the function v(x) 

ILATE rule is Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

Calculation:

I = ∫ ex cos x dx

Integrating by ILATE rule,

I = cos x ∫ e^x dx - ∫ (-sin x ∫ e^x dx) dx

I = e^x cos x + ∫ e^x sin x dx

I = ex cos x + sin x ∫ ex dx - ∫ (cos x ∫ e^x dx) dx

I = ex cos x + e^x sin x - ∫ ex cos x dx

I = ex cos x + e^x sin x - I

2I = ex cos x + e^x sin x

I = \(\boldsymbol{\rm {e^x \cos x + e^x \sin x\over 2}}\)

Concept:

• \(\smallint {e^{ax}}\;dx = \frac{{{e^{ax}}}}{a} + C\) where C is a constant

Integration by parts:

The formula for integrating by parts is given by;

\( \Rightarrow {\rm{}}\smallint {\rm{u\;vdx}} = {\rm{u\;}}\smallint {\rm{vdx}} - {\rm{\;}}\smallint \left( {\frac{{{\rm{du}}}}{{{\rm{dx}}}}{\rm{\;}}\smallint {\rm{vdx}}} \right){\rm{dx}}\)

Where u is the function u(x) and v is the function v(x)

• ILATE Rule: Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

Calculation:

Here, we have to find the value of \(\smallint {x}{e^{3x}}\;dx\)

According to the integration by parts (ILATE Rule) we have: x as our first function and e^3x as our second function

i.e u(x) = x and v(x) = e^3x

As we know that, \({\rm{}}\smallint {\rm{u\;vdx}} = {\rm{u\;}}\smallint {\rm{vdx}} - {\rm{\;}}\smallint \left( {\frac{{{\rm{du}}}}{{{\rm{dx}}}}{\rm{\;}}\smallint {\rm{vdx}}} \right){\rm{dx}}\)

\(\Rightarrow \smallint x{e^{3x}}\;dx = x \cdot \smallint {e^{3x}}\;dx - \;\smallint \left\{ {\frac{{d\left( x \right)}}{{dx}} \cdot \;\smallint {e^{3x}}\;dx} \right\}dx\)

As we know that, \(\smallint {e^{ax}}\;dx = \frac{{{e^{ax}}}}{a} + C\) where C is a constant

\(\Rightarrow \smallint x{e^{3x}}\;dx = x \cdot \frac{{{e^{3x}}}}{3} - \;\smallint \left[ {1 \cdot \frac{{{e^{3x}}}}{3}} \right]dx\)

As we know that, \(\smallint {e^{ax}}\;dx = \frac{{{e^{ax}}}}{a} + C\) where C is a constant

\(\Rightarrow \smallint x{e^{3x}}\;dx = \frac{{{e^{3x}} \cdot \left( {3x - 1} \right)}}{9} + C\)

Concept:

Greatest Integer Function [X]: It indicates an integral part of the real number x  which is the nearest and smaller integer to x. It is also known as the floor of X.
[x] = The largest integer that is less than or equal to x.

Calculation:

\(Let,\ \ I\ = \ \int_0^{1.5} {\left[ {{x^2}} \right]} dx\)

\(⇒ \ I\ = \int_0^1 [x^2]dx +\int_1^{√{2}} [x^2] dx + \int_{√{2}}^{1.5} [x^2]dx\)

\(⇒ \ I\ =\displaystyle\int_0^1 0 \cdot dx + \int_1^{√{2}} 1 \cdot dx + \int_{√{2}}^{1.5} 2 \cdot dx\)

\(⇒ \ I\ =0 + (√{2}-1)+2 (1.5 -√{2}) \)

⇒ I = 2 - √2

Hence, the value of the integral is  2 - √2.

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Integration by parts: Integration by parts is a method to find integrals of products. The formula for integrating by parts is given by:

⇒ \(\rm ∫ u vdx=u ∫ vdx- ∫ \left({du\over dx}\times \int vdx\right)dx \) + C

where u is the function u(x) and v is the function v(x) 

ILATE rule is Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

Calculation:

I = ∫ (x ln x) dx

Using integration by parts

⇒ I = ln x ∫ x dx - ∫ \(\rm \left({d\ln x\over dx}\times \int xdx\right)\) dx + C

⇒ I = \(\rm {x^2\over2}\) ln x - ∫ \(\rm \left({1\over x}\times{x^2\over2}\right)\) dx + C

⇒ I = \(\rm {x^2\over2}\) ln x - ∫ \(\rm {x\over2}\) dx + C

⇒ I = \(\boldsymbol{\rm {x^2\ln x\over2}-}\) \(\boldsymbol{\rm {x^2\over4}}\) + C

Given:

The given integral is \(\int_0^3 {\left| {1 - x} \right|} \,{\rm{dx}}\)

Formula Used:

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

Calculation:

We have,

⇒ I = \(\int_0^3 {\left| {1 - x} \right|} \,{\rm{dx}}\)

Now |1 - x| = 1 - x  for 1 - x ≥ 0 ⇒1 ≥ x and   

 |1 - x| = - (1 - x)  for 1 - x < 0 ⇒1 < x

⇒ I = \(\int_0^1 {\left| {1 - x} \right|} \,{\rm{dx}} + \int_1^3 {\left| {1 - x} \right|} \,{\rm{dx}}\)

⇒ I = \(\int_0^1 { {(1 - x)} } \,{\rm{dx}} \ - \ \int_1^3 { {(1 - x)} } \,{\rm{dx}}\)

⇒ I = \(\left[ x - \frac{x^2}{2} \right]_0^1 - \left[x - \frac{x^2}{2} \right]_1^3\)

⇒ I = \(1 - \frac{1}{2} - \left[ 3 - \frac{9}{2} - 1 + \frac{1}{2} \right]\)

⇒ I = \(1 - \frac{1}{2} - 3 + \frac{9}{2} + 1 - \frac{1}{2}\)

⇒ I = \(2 - 1 - 3 + \frac{9}{2}\)

⇒ I = \(- 2 + \frac{9}{2}\)

⇒ I = \(\frac{5}{2}\)

∴ \(1 - \frac{1}{2} - \left[ 3 - \frac{9}{2} - 1 + \frac{1}{2} \right]\) equals to \(\frac{5}{2}\)

Explanation:

Given Integral is,

\(\displaystyle∫_0^1 \dfrac{e^t}{1+t}dt=a\)

Considering \(I_1 ~=~\frac{1}{1+t}\) and \(I_2 ~=~e^t \)

Hence by Integration by parts we know that,

∫ [I₁][ I₂ ] dt = I₁∫ I₂ dt - ∫ [I'₁∫ I₂ dt]dt

\(\frac{1}{1+t}\int_0^1 e^t dt-\int_0 ^1 [\frac{d}{dt}(\frac{1}{1+t}).\int_0 ^1 e^t dt]dt~=~a\)

\(\frac{1}{1+t}[e^t ]_0^1-\int_0^1 (\frac{-1}{(1+t)^2}).e^t dt~=~a\)

\([\frac{e^t}{1+t}]_0^1 +\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a\)

\(\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a-[\frac{e^t}{1+t}]_0^1\)

\(\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a-[\frac{e^1}{1+1}-\frac{e^0}{1+0}]\)

\(\int_0^1 \frac{e^t}{(1+t)^2}dt~=~a-[\frac{e}{2}-1]\)

\(\displaystyle\int_0^1 \dfrac{e^t}{(1+t)^2}dt~=~a+1-\frac{e}{2}\)

Concept:

1. Integration by parts: 

Integration by parts is a method to find integrals of products

The formula for integrating by parts is given by;

\(⇒ \smallint {\rm{u\;vdx}} = {\rm{u\;}}\smallint {\rm{vdx}} - {\rm{\;}}\smallint \left( {\frac{{{\rm{du}}}}{{{\rm{dx}}}}{\rm{\;}}\smallint {\rm{vdx}}} \right){\rm{dx}}\)

Where u is the function u(x) and v is the function v(x)

2. ILATE Rule: 

Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

Calculation:

Let I = ∫ (e^{log x} + sin x) cos x dx

\(⇒\rm \int (x \cos x + \sin x \cos x) dx\)        (∵ e^{log x} = x)

\(⇒ \rm \int x\cos x \;dx + \int \sin x \cos x\;dx\)

⇒ I = I₁ + I₂               .... (1)

Now, I₁ = \(\rm \int x\cos x \;dx\)

Applying by parts, we get

\(⇒ {\rm{x\;}}\smallint {\rm{\cos x\;dx}} - {\rm{\;}}\smallint \left( {\frac{{{\rm{dx}}}}{{{\rm{dx}}}}{\rm{\;}}\smallint {\rm{\cos x\;dx}}} \right){\rm{dx}} \)

⇒ x sin x - \(\rm \int \sin x \;dx \) + c

⇒ x sin x + cos x + c

Now, I₂ = \(\rm \int \sin x \cos x\;dx\)

Let sin x = t

Differentiating with respect to x, we get

⇒ cos x dx = dt

⇒ I₂ = \(\rm \int tdt = \frac{t^2}{2} + c\)

⇒ \(\rm \frac{\sin^2 x}{2} + c\)

Put the value of I₁ and I₂ in equation (1), we get

∴ The required integral (I) is \(\rm x \sin x + \cos x + \dfrac{\sin^2 x}{2}+c\).

Concept:

• \(\smallint {e^x} \cdot \left[ {f\left( x \right) + f'\left( x \right)} \right]\;dx = {e^x} \cdot f\left( x \right) + C\) where C is a constant
• \(\frac{{d\left( {{x^n}} \right)}}{{dx}} = n \cdot {x^{n - 1}}\)

Calculation:

Here we have to find the value of \(\smallint \frac{{x \cdot {e^x}}}{{{{\left( {1 + x} \right)}^2}}}dx\)

Now the given integral can be re-written as:

\(\Rightarrow \smallint \frac{{x \cdot {e^x}}}{{{{\left( {1 + x} \right)}^2}}}dx = \;\smallint {e^x} \cdot \left\{ {\frac{{\left( {1 + x} \right) - 1}}{{{{\left( {1 + x} \right)}^2}}}} \right\}dx\)

\(\Rightarrow \smallint {e^x} \cdot \left\{ {\frac{{\left( {1 + x} \right) - 1}}{{{{\left( {1 + x} \right)}^2}}}} \right\}dx = \;\smallint {e^x} \cdot \left\{ {\frac{1}{{1 + x}} - \frac{1}{{{{\left( {1 + x} \right)}^2}}}} \right\} dx\)

Now by comparing the above equation with \(\smallint {e^x} \cdot \left[ {f\left( x \right) + f'\left( x \right)} \right]\;dx\) we get

\(f\left( x \right) = \;\frac{1}{{1 + x}}\;\;and\;f'\left( x \right) = \;\frac{1}{{{{\left( {1 + x} \right)}^2}}}\)

As we know that, \(\smallint {e^x} \cdot \left[ {f\left( x \right) + f'\left( x \right)} \right]\;dx = {e^x} \cdot f\left( x \right) + C\) where C is a constant

\(\Rightarrow \smallint \frac{{x \cdot {e^x}}}{{{{\left( {1 + x} \right)}^2}}}dx = \frac{{{e^x}}}{{1 + x}} + C\) where C is a constant

Concept:

Integration by parts

Integration by parts is used to integrate the product of two or more functions. The two functions

to be integrated f(x) and g(x) are of the form \(\rm \displaystyle\int \)f(x).g(x). Thus, it can be called a product rule of

integration. Among the two functions, the first function f(x) is selected such that its derivative

formula exists, and the second function g(x) is chosen such that an integral of such a function exists.

\(\rm \displaystyle\int \)f(x).g(x).dx = f(x)\(\rm \displaystyle\int \)g(x).dx−\(\rm \displaystyle\int \big[\)f′(x)\(\rm \displaystyle\int \)g(x).dx\(\big]\).dx + C

A useful rule of integral by parts is ILATE.

I: Inverse trigonometric functions : sin^-1(x), cos^-1(x), tan^-1(x)

L: Logarithmic functions : ln(x), log(x)

A: Algebraic functions : x², x³

T: Trigonometric functions : sin(x), cos(x), tan (x)

E: Exponential functions : e^x, 3^x

Properties of logarithms

\(\rm \displaystyle e^{\log_e x}\) = x

Let y = \(\rm \displaystyle e^{\log x}\)

Taking log both sides,

⇒ log y = log \(\rm \displaystyle e^{\log x}\)

⇒ log y = log x log e

⇒ log y = log x

⇒ y = x

Calculation:

I = \(\rm \displaystyle\int e^{\log x} \sin x \ dx\) 

⇒ I = \(\rm \displaystyle\int \)x sin x dx

Taking x as the first function and sin x as the second function and integrating by parts, we get,

⇒ I = x\(\rm \displaystyle\int \)sin x dx − \(\rm \displaystyle\int \big[\big(\frac{d(x)}{dx}\big) \)\(\rm \displaystyle\int \)sin x dx\(\big]\)dx

⇒ I = x (− cos x) − \(\rm \displaystyle\int \)1.(−cos x) dx

⇒ I = −x cos x + sin x + C

∴ The value of \(\rm \displaystyle\int e^{\log x} \sin x \ dx\) is (sin x - x cos x) + C