Evaluate using Integration by Parts MCQ Quiz - Objective Question with Answer for Evaluate using Integration by Parts - Download Free PDF

Explanation:

Given:

f(x) =|x² – x – 2|

= {x²- x - 2; x ∈ (-∞, -1) ∪ (2, ∞) 

      - (x2 -x -2 ; x ∈[ -1,2]

Let I = \(\int_1^3f(x)dx\)

= \(- \int_1^2(x^2 -x-2)dx+ \int_3^2(x^2-x-2)dx\)

= \(- [\frac{x^3}{3} -\frac{x^2}{2}-2x]^2_1 + [\frac{x^3}{3} -\frac{x^2}{2}-2x]^3_2\)

= \([\frac{8}{3}-\frac{4}{2}-4] + [[\frac{1}{3}-\frac{1}{2}-2] + [\frac{27}{3}-\frac{9}{2}-6] - [\frac{8}{3}-\frac{4}{2}-4] \)

= \(\frac{20}{6} - \frac{13}{6} -\frac{9}{6} +\frac{20}{6}\) = 3

∴ Option (b) is correct.

Explanation:

Given:

f(x) =|x² – x – 2|.

= {x² – x – 2; x ∈ (-∞ –1) ∪  (2,∞ ) – (x² – x – 2); x ∈ [–1, 2]

Let I = - \(\int_0^2(x^2 – x – 2)dx\)

= - \([\frac{x^3}{3}-\frac{x^2}{2} -2x]_2^0\)

= \(- [\frac{8}{3}-\frac{4}{2}+4] +0 =\frac{10}{3}\)

∴ Option (d) is correct

Calculation

\(\int f(x)g(x)dx=f(x) \int g(x)dx=\int \left(\dfrac{d}{dx} [f(x)] \int g(x)dx \right)dx\)

\(\int_0^1 4\underset{I}{x^3} \frac{d^2}{dx^2} (1-\underset{II}{x^2})^5 dx\)

\(= \left [ 4x^3 \frac{d}{dx} (1-x^2)^5 \right ]_0^1 - \int_0^1 12x^2 \frac{d}{dx} (1 - x^2)^5 dx\)

\(=\left [ 4x^3 \times 5(1-x^2)^4 (-2x) \right ]_0^1 - 12 \left [ \left [ x^2 (1-x^2)^5 \right ]_0^1 - \int_0^1 2x (1-x^2)^5 dx \right ]\)

\(=0-0-12 [0-0] + 12 \int_0^1 2x (1-x^2)^5 dx\)

\(=12 \times \left [ - \frac{(1-x^2)^6}{6} \right ]_0^1\)

\(=12 \left [ 0 + \dfrac{1}{6} \right ] \) = 2

Explanation -

Given the equation:

\(3f(x) + f\left(\frac{1}{x}\right) = \frac{1}{x} + 1\) ..... (i)

Substitute x with 1/x :

\(3f\left(\frac{1}{x}\right) + f(x) = x + 1\)...... (ii)   

We now have two equations:

\(3f(x) + f\left(\frac{1}{x}\right) = \frac{1}{x} + 1\)

\(3f\left(\frac{1}{x}\right) + f(x) = x + 1\)

Let: f(x) = a and f(1/x) = b

The system of equations becomes:

3a + b = 1/x + 1
3b + a = x + 1

Multiply the first equation by 3:

9a + 3b = 3/x + 3

Subtract the second equation from the modified first equation:

9a + 3b - (3b + a) = 3/x + 3 - (x + 1)

⇒ 8a = 3/x + 3 - x - 1

⇒ 8a = 3/x - x + 2

⇒ a =\( \frac{1}{8} \left(\frac{3}{x} - x + 2\right)\)

Therefore: f(x) = \( \frac{1}{8} \left(\frac{3}{x} - x + 2\right)\)  

Thus, the function f(x) = \(\frac{3}{8x} - \frac{x}{8} + \frac{1}{4}\)

Now  \(8 \int_1^2 f(x) d x\)

= \(8 \int_1^2 [ \frac{3}{8x} - \frac{x}{8} + \frac{1}{4} ]d x\)

= \(8 [ \frac{3}{8}ln (x) - \frac{x^2}{16} + \frac{x}{4} ]_1^2 \)

= \( [ 3ln (x) - \frac{x^2}{2} + 2x ]_1^2 \)

= \( [ 3ln (2) - \frac{4}{2} + 4 ] - [ - \frac{1}{2} + 2]\)

= \( [ 3ln (2) + 2 ] - \frac{3}{2}\)

= \( 3ln (2) + \frac{1}{2}\)

= \( 3ln (2) + \frac{1}{2}ln e\)

= ln 8 + ln√e

= ln 8√e

Hence Option (1) is Correct.

Explanation -

Given the equation:

\(3f(x) + f\left(\frac{1}{x}\right) = \frac{1}{x} + 1\) ..... (i)

Substitute x with 1/x :

\(3f\left(\frac{1}{x}\right) + f(x) = x + 1\)...... (ii)   

We now have two equations:

\(3f(x) + f\left(\frac{1}{x}\right) = \frac{1}{x} + 1\)

\(3f\left(\frac{1}{x}\right) + f(x) = x + 1\)

Let: f(x) = a and f(1/x) = b

The system of equations becomes:

3a + b = 1/x + 1
3b + a = x + 1

Multiply the first equation by 3:

9a + 3b = 3/x + 3

Subtract the second equation from the modified first equation:

9a + 3b - (3b + a) = 3/x + 3 - (x + 1)

⇒ 8a = 3/x + 3 - x - 1

⇒ 8a = 3/x - x + 2

⇒ a =\( \frac{1}{8} \left(\frac{3}{x} - x + 2\right)\)

Therefore: f(x) = \( \frac{1}{8} \left(\frac{3}{x} - x + 2\right)\)  

Thus, the function f(x) = \(\frac{3}{8x} - \frac{x}{8} + \frac{1}{4}\)  

Concept:

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

The formula for integrating by parts is given by;

⇒ ∫ uv dx = u(x) ∫ v(x) dx - ∫ [u'(x) ∫ v(x) dx] dx

Where u is the function u(x) and v is the function v(x), and is chosen by following ILATE rule.

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

Calculation:

Given:

Let I = \(\displaystyle∫_0^1 \rm xe^x\ dx\)

Applying by parts rule, we get

I \(\rm = x∫ _0^1 e^{x}\;dx - ∫ _0^1(\frac{dx}{dx}∫ e^{x}\;dx)\;dx \)

\(I\rm = [xe^{x}]_{0}^{1} - ∫ _0^11\times e^{x}\;dx\)

\(I\rm = [xe^{x}]_{0}^{1} - [e^{x}]_{0}^{1} \)

I = (e - 0) - (e - 1)

I = 1

Hence the required value of integration is 1.

Concept:

Modulus Function:

A modulus function is a function which gives the absolute value of a number or variable.

\(f(x)= |x| = \left\{ {\begin{array}{*{20}{c}} {x \ if \ x \ge 0}\\ { - x} \ if x < 0 \end{array}} \right.\)

Formula Used:

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

Calculation:

We have,

⇒ I = \(\int\limits_0^2 {|1-x|}dx\)

⇒ 1 - x < 0 for 1 < x ≤ 2 and 1 - x ≥ 0 for 0 < x ≤ 1

⇒ I = \(\int\limits_0^1(1 - x)dx \ - \ \int\limits_1^2 (1 - x)dx\)

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

⇒ I = \(\left[ 1- \frac{1}{2} - 0\right]\) - \(\left[ 2 - \frac{2^2}{2} - 1 + \frac{1}{2} \right]\)

⇒ I = \(\left( \frac{1}{2} \right) - \left( 0 - \frac{1}{2} \right)\)

⇒ I = \(\frac{1}{2} + \frac{1}{2}\)

⇒ I = 1

∴ The value of \(\int\limits_0^2 {|1-x|}dx\) is 1.

Concept:

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

• The formula for integrating by parts is given by,
• ∫u v dx = u∫v dx −∫u' (∫v dx) 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:

Let I = \(\int^\pi_0 x^3 \sin xdx\)

Apply by parts rule, we get

\(\rm =x^3 \int^\pi_0sinxdx- \int^\pi_03x^2(-cosx)dx\)

\(\rm =[x^3(-cosx)]_0^\pi+3[x^2\int^\pi_0cosxdx- \int^\pi_02x(sinx)dx]_0^\pi\)

\(\rm =\pi^3+0-6\int^\pi_0x(sinx)dx\)

\(\rm=\pi^3-6[x\int^\pi_0sinxdx- \int^\pi_0(-cosx)dx]\)

\(\rm =\pi^3-6[\pi- 0]\)

= π3 - 6π

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;

 \(\Rightarrow \smallint {\rm{u\;vdx}} = {\rm{u\;}}\smallint {\rm{vdx}} - {\rm{\;}}\smallint \left( {\frac{{{\rm{du}}}}{{{\rm{dx}}}}{\rm{\;}}\smallint {\rm{vdx}}} \right){\rm{dx}}\), 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 = \(\rm \displaystyle\int_0^1 x \tan^{-1} xdx \)

Apply by parts rule,

\(\rm= [{\tan ^{ - 1}}x \cdot \;\displaystyle\int x\;dx]_0^1 - \;\displaystyle\int_0^1 \left\{ {\frac{{d\left( {{{\tan }^{ - 1}}x} \right)}}{{dx}} \cdot \;\displaystyle\int x\;dx} \right\}dx\)

\( = \rm [{\tan ^{ - 1}}x \cdot \frac{{{x^2}}}{2}]_0^1 - \;\displaystyle\int_0^1 \frac{1}{{1 + {x^2}}} \cdot \frac{{{x^2}}}{2}\;dx\)

\( = \rm [{\tan ^{ - 1}}x \cdot \frac{{{x^2}}}{2}]_0^1 - \;\dfrac{1}{2}\displaystyle\int_0^1 \frac{1+x^2-1}{{1 + {x^2}}}\;dx\)

\( = \rm [{\tan ^{ - 1}}x \cdot \frac{{{x^2}}}{2}]_0^1 - \;\dfrac{1}{2}\displaystyle\int_0^1 [1 -\frac{1}{{1 + {x^2}}}]\;dx\)

\( = \rm [{\tan ^{ - 1}}x \cdot \frac{{{x^2}}}{2}]_0^1 - \frac{1}{2} [x - {\tan ^{ - 1}}x ]_0^1\)

\( = \rm [{\tan ^{ - 1}}1 \cdot \frac{{{1^2}}}{2}] - \frac{1}{2} [1 - {\tan ^{ - 1}}1 ]\\= \dfrac{\pi}{4}\times \dfrac{1}{2} - \dfrac{1}{2}+\dfrac{1}{2} \times \dfrac{\pi}{4}\)

= \(\rm \dfrac{\pi}{4}-\dfrac{1}{2}\)

Concept:

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

The formula for integrating by parts is given by;

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

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:

First we will calculate the integration without the limits.

Let us consider \(\rm \cot^{-1}x=u\) .

On differentiating u on both sides we get,

 \(\rm -\dfrac{1}{1+x^2}dx=du\) .

Therefore, we integrate the given function as follows:

\(\rm \int cot^{-1}x\,dx = x\cot^{-1}x + \int \dfrac{x}{1+x^2}\, dx\\ = x\cot^{-1}x + \dfrac{1}{2}\ln(1+x^2) +C \)

Now as the given integral is definite we will remove the constant of integration and put the limits.

\(\begin{align*} \int_{-3}^{3}\cot^{-1}x &= \left[x\cot^{-1}x+\dfrac{1}{2}\ln(1+x^2)\right]_{-3}^{3}\\ &= \left[3\cot^{-1}3+\dfrac{1}{2}\ln(10)\right]-\left[-3\cot^{-1}(-3)+\dfrac{1}{2}\ln(10)\right]\\ &= 3\left(\cot^{-1}(3)+\cot^{-1}(-3)\right)\\ &= 3\left(\cot^{-1}(3)+\pi -\cot^{-1}(3)\right)\\ &= 3\pi \end{align*}\)

Therefore, \(\displaystyle\int_{-3}^{3}\cot^{-1}x\, dx = 3\pi\).

Concept:

\(\smallint f\left( x \right)g\left( x \right){\bf{dx}} = {\bf{f}}\left( {\bf{x}} \right)\smallint g\left( x \right)\;dx - \smallint \left[ {\left( {\frac{d}{{dx}}f\left( x \right)} \right)\smallint g\left( x \right)dx} \right]dx\)

Calculation:

Given:

\(\smallint x{e^{x\;}}dx = x\smallint {e^x}\;dx - \smallint \left[ {\left( {\frac{d}{{dx}}x} \right)\smallint {e^x}\;dx} \right]dx\)

\( = {\rm{x}}{{\rm{e}}^{\rm{x}}} - \smallint 1{\rm{\;}}{{\rm{e}}^{\rm{x}}}{\rm{\;dx}}\)

\( = {\rm{x}}{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{\rm{x}}}{\rm{\;}}\)

Applying limits,

\(\mathop \smallint \limits_0^1 {\rm{x}}{{\rm{e}}^{\rm{x}}}{\rm{\;dx}} = \left[ {{\rm{x}}{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{\rm{x}}}} \right]_0^1\)

= [1 × e¹ – e¹] – [0 × e⁰ – e⁰]

= [e¹ – e¹] – [0 - 1]

= 1

Explanation:

Given:

f(x) =|x² – x – 2|.

= {x² – x – 2; x ∈ (-∞ –1) ∪  (2,∞ ) – (x² – x – 2); x ∈ [–1, 2]

Let I = - \(\int_0^2(x^2 – x – 2)dx\)

= - \([\frac{x^3}{3}-\frac{x^2}{2} -2x]_2^0\)

= \(- [\frac{8}{3}-\frac{4}{2}+4] +0 =\frac{10}{3}\)

∴ Option (d) is correct

Explanation:

Given:

f(x) =|x² – x – 2|

= {x²- x - 2; x ∈ (-∞, -1) ∪ (2, ∞) 

      - (x2 -x -2 ; x ∈[ -1,2]

Let I = \(\int_1^3f(x)dx\)

= \(- \int_1^2(x^2 -x-2)dx+ \int_3^2(x^2-x-2)dx\)

= \(- [\frac{x^3}{3} -\frac{x^2}{2}-2x]^2_1 + [\frac{x^3}{3} -\frac{x^2}{2}-2x]^3_2\)

= \([\frac{8}{3}-\frac{4}{2}-4] + [[\frac{1}{3}-\frac{1}{2}-2] + [\frac{27}{3}-\frac{9}{2}-6] - [\frac{8}{3}-\frac{4}{2}-4] \)

= \(\frac{20}{6} - \frac{13}{6} -\frac{9}{6} +\frac{20}{6}\) = 3

∴ Option (b) is correct.

Concept:

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

The formula for integrating by parts is given by;

⇒ ∫ uv dx = u(x) ∫ v(x) dx - ∫ [u'(x) ∫ v(x) dx] dx

Where u is the function u(x) and v is the function v(x), and is chosen by following ILATE rule.

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

Calculation:

Given:

Let I = \(\displaystyle∫_0^1 \rm xe^x\ dx\)

Applying by parts rule, we get

I \(\rm = x∫ _0^1 e^{x}\;dx - ∫ _0^1(\frac{dx}{dx}∫ e^{x}\;dx)\;dx \)

\(I\rm = [xe^{x}]_{0}^{1} - ∫ _0^11\times e^{x}\;dx\)

\(I\rm = [xe^{x}]_{0}^{1} - [e^{x}]_{0}^{1} \)

I = (e - 0) - (e - 1)

I = 1

Hence the required value of integration is 1.