Properties of Definite Integrals MCQ Quiz in मल्याळम - Objective Question with Answer for Properties of Definite Integrals - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Some useful formulas are:

\(\rm {d\over {dx}}{(cosx)}=-sinx\)

\(\rm \int {1\over {(1+x^2)}}dx=tan^{-1}x+c\)

tan^-11 = \(\pi \over 4\)

tan^-10 = 0

\(\rm ∫_p^q f(x) dx = – ∫_q^p f(x)dx\)

Calculation:

\(\rm \int_0^{π \over2} {sinx\over {(1+cos^2x)}}dx\)

Let, cosx = z

∴ - sinx dx = dz 

Substituting the values we get,

\(\rm \int_1^{0}- {1\over {(1+z^2)}}dz\)

= \(\rm \int_0^{1}{1\over {(1+z^2)}}dz\)

= \(\rm tan^{-1}z|_0^1\)

= tan^-11 - tan^-10

= \(\pi \over 4\)

Concept:

For an odd function f(x): \(\rm \int_{-a}^{\ \ a}f(x)\ dx=0\).

Calculation:

We observe that ax⁵ and bx³ are odd functions of x, 

∵ a(-x)5 = -ax⁵ and b(-x)3 = -bx3.

∴ \(\rm \int_{-2}^{\ \ 2}ax^5\ dx=0\) and \(\rm \int_{-2}^{\ \ 2}bx^3\ dx=0\).

The value of \(\rm \int_{-2}^{\ \ 2}(ax^5 + bx^3 + c)\ dx\) depends on the value of c.

In fact, 

\(\rm \int_{-2}^{\ \ 2}(ax^5 + bx^3 + c)\ dx=\int_{-2}^{\ \ 2}c\ dx\).

= c [2 - (-2)]

= 4c

Additional Information

A function f(x) is:

• Even, if f(-x) = f(x). And \(\rm \int_{-a}^{\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).
• Odd, if f(-x) = -f(x). And \(\rm \int_{-a}^{\ \ a}f(x)\ dx=0\).

Concept:

Definite Integral properties:

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)

Trigonometry Formula:

tan ( -x ) = - tan x 

Calculation:

Let , I = \(\rm \int_{-π/2}^{π/2}\frac{1}{1+e^{\tan x}}dx\)                 ....(1)

Now using property, \(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)

I = \(\rm \int_{-π/2}^{π/2}\frac{1}{1+e^{\tan (-x)}}dx\)

I = \(\rm \int_{-π/2}^{π/2}\frac{1}{1+e^{-\tan x}}dx\)                    .... (ii)

Adding equation (i) and (ii), we get

2I = \(\rm\int_{-π/2}^{π/2}\left \{ \frac{1}{1+e^{tanx}} +\frac{1}{1+e^{-tanx}}\right \}dx\)  

2I = \(\rm\int_{-π/2}^{π/2}\left \{ \frac{1}{1+e^{tanx}} +\frac{e^{tanx}}{1+e^{tanx}}\right \}dx\) 

2I = \(\rm\int_{-π/2}^{π/2}\frac{1+e^{tanx}}{1+e^{tanx}}dx\) 

2I = \(\rm\int_{-π/2}^{π/2}1 dx\) 

2I = π

∴  I = π /2  

The correct option is 3. 

Concept:

Integral Properties:

• \(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{dx}}\)

Calculation:

Given: f(x) = f(a – x) & g(x) + g(a – x) = 2

To find: \(\mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right){\rm{dx}}\)

\({\rm{I}} = \mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right){\rm{dx}}\)      …. (1)

Using integral property, \(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{dx}}\)

\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{g}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{dx}}\)

Using given, f(x) = f(a – x)

\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{g}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{dx}}\)      …. (2)

Adding equation (1) & (2)

\(2{\rm{I}} = \mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{g}}\left( {\rm{x}} \right){\rm{dx}} + \mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{g}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{dx}}\)

\(\Rightarrow 2{\rm{I}} = \mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right)\left\{ {{\rm{g}}\left( {\rm{x}} \right) + {\rm{g}}\left( {{\rm{a}} - {\rm{x}}} \right)} \right\}{\rm{dx}}\)

Using given, g(x) + g(a – x) = 2

\( \Rightarrow 2{\rm{I}} = \mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right) \times 2{\rm{dx}}\)

\(\Rightarrow {\rm{I}} = \mathop \smallint \limits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}}\)

Concept:

Some useful formulas are:

\(\rm \int_a^b f(x) dx = \int_a^b f(a + b – x) dx\)

\(\rm cot^{-1}(-x)=(π -cot^{-1}x) \ dx\)

\(\rm ∫ x^n dx = \frac {(x^{n+1})} {(n\;+\;1)} +C; \ n≠1\)

Calculation:

Let, I = \(\rm \int_{-π}^{π} cot^{-1}x \ dx\)

Since, \(\rm \int_a^b f(x) dx = \int_a^b f(a + b – x) dx\) , so

 I = \(\rm \int_{-π}^{π} cot^{-1}({-π +π -x} )\ dx\)

∴ I = \(\rm \int_{-π}^{π} cot^{-1}(-x) \ dx\)

∴ I = \(\rm \int_{-π}^{π} (π -cot^{-1}x) \ dx\)

∴ I = \(\rm \int_{-π}^{π} π \ dx\) - \(\rm \int_{-π}^{π} cot^{-1}x \ dx\)

∴  I = \(\rm \int_{-π}^{π} π \ dx\) - I

∴ 2I = \(\rm π x|_{-π}^π = π[ {π - (-π)}]= 2π^2\)

∴ I = π² 

Concept: 

The rules for integrating even and odd functions ,

If the function is even or odd and the interval is [-a, a], we can apply these rules:

• When f(x) is even ⇔ \(\rm\int_{-a}^{a}f(x)dx= 2 \int_{0}^{a}f(x)dx\)
• When f(x) is odd ⇔ \(\rm\int_{-a}^{a}f(x)dx= 0\) 

​Calculation: 

Let f (x) = x⁵ sin⁴ x 

f (-x) = (-x)⁵ sin⁴ (-x) = -x⁵ (-sin x )⁴ = -x⁵ sin⁴ x .

⇒ f(-x) = - f(x) 

So, f(x) is an odd function . 

We knoiw that, when f(x) is odd ⇔ \(\rm\int_{-a}^{a}f(x)dx= 0\) 

Hence , \(\rm \int_{-\pi/2}^{\pi/2} x^{5}\sin^{4}x dx\) = 0 .  

The correct option is 4 .

Calculation:

Given: \(\rm \displaystyle\int_0^{\pi/2}\log \cos x \ dx=\dfrac{\pi}{2}\log \left(\dfrac{1}{2}\right)\)          .... (1)

Now,

\(\text {Let}\; \rm I = \rm \displaystyle\int_0^{\pi/2}\log \sec x \ dx \\=\rm \displaystyle\int_0^{\pi/2}\log (\frac{1}{\cos x}) \ dx\\\)

\(= \rm \displaystyle\int_0^{\pi/2}\log 1 \ dx -\rm \displaystyle\int_0^{\pi/2}\log \cos x \ dx \)        (∵ log \(\rm \frac{m}{n} = \log m - \log n)\)

\(= 0-\rm \displaystyle\int_0^{\pi/2}\log \cos x \ dx \)                            (∵ log 1 = 0)

From equation (1), we get

\(\rm I = -\dfrac{\pi}{2}\log (\dfrac{1}{2})\\= -\dfrac{\pi}{2}[\log1 - \log 2]\)

= \(\dfrac{\pi}{2}\log 2\)

Concept:

Property of definite integrals: 

\(\mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{dx}}\)

Calculation:

I = \(\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\left( {3\sin {\rm{x}} + 4\cos {\rm{x}}} \right)}}{{\sin {\rm{x}} + \cos {\rm{x}}}}{\rm{dx}}\)         …. (1)

Using the property of definite integrals: 

\(\mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{dx}}\)

I \( = \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\left[ {3\sin \left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right) + 4\cos \left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)} \right]}}{{\sin \left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right) + \cos \left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)}}{\rm{dx}}\)

I \( = {\rm{\;}}\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\left( {3\cos {\rm{x}} + 4\sin {\rm{x}}} \right)}}{{\cos {\rm{x}} + \sin {\rm{x}}}}{\rm{dx}}\)       .... (2)

Adding equations (1) and (2), we get

\(2{\rm{I}} = \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\left( {3\sin {\rm{x}} + 4\cos {\rm{x}}} \right)}}{{\sin {\rm{x}} + \cos {\rm{x}}}}{\rm{dx}} + \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\left( {3\cos {\rm{x}} + 4\sin {\rm{x}}} \right)}}{{\cos {\rm{x}} + \sin {\rm{x}}}}{\rm{dx}} \)

\(= {\rm{\;}}\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\left( {7\cos {\rm{x}} + 7\sin {\rm{x}}} \right)}}{{\cos {\rm{x}} + \sin {\rm{x}}}}{\rm{dx}}\)

\( = {\rm{\;}}\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{7\left( {\cos {\rm{x}} + \sin {\rm{x}}} \right)}}{{\cos {\rm{x}} + \sin {\rm{x}}}}{\rm{dx}} \)

\(= {\rm{\;}}\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} 7{\rm{dx}} \)

\(= {\rm{\;}}\left[ {7x} \right]_0^{\frac{{\rm{\pi }}}{2}}\)

\(= 7\left( {\frac{{\rm{\pi }}}{2} - 0} \right) \)

\(= \frac{{7{\rm{\pi }}}}{2}\)

∴ I = \( \frac{{7{\rm{\pi }}}}{4}\)

Concept:

Odd and even function:

If a function f(x) such that f(-x) = f(x) then f(x) is an even function and if f(-x) = -f(x) then f(x) is an odd function.

Integration of odd and even function:

The following two cases hold for a differentiable function f(x).

If f(x) is an even function then \( \rm \displaystyle\int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx\).

If f(x) is an odd function then \(\rm \displaystyle\int_{-a}^{a}f(x)dx = 0\).

Calculation:

Let \(\rm f(x) = x\sin x\), put -x instead of x then we get \(\rm f(-x) = (-x)\sin(-x) = x\sin x\).

Therefore, f(-x) = f(x) implies that f(x) is an even function.

Thus, the integral is given by,

\(\rm \begin{align*} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x\sin x dx &= 2\int_{0}^{\frac{\pi}{2}}x\sin x dx\\ &= 2\left[x(-\cos x) - \int(-\cos xdx)\right]_{0}^{\frac{\pi}{2}}\\ &= 2\left[-x\cos x + \sin x\right]_{0}^{\frac{\pi}{2}}\\ &= 2\left[\left(-\dfrac{\pi}{2}\cos\left(\dfrac{\pi}{2}\right) + \sin\left(\dfrac{\pi}{2}\right)\right)-\left(-0\cos 0+\sin 0\right)\right]\\ &= 2 \end{align*}\)

Hence, the value of the given integral is 2.

Concept:

\(\rm \int_{0}^{2a} f(x) d x=2\int_{0}^{a} f(2a-x) d x\)

Calculation:

\({\rm{I}} = \mathop \smallint \limits_0^{\rm{\pi }} {\rm{ln\;}}(\sin {\rm{x}}){\rm{\;dx}}\)

\(\begin{array}{l} I=2\int_{0}^{\pi/2 } \ln (\sin (\pi-x) d x \\ =2\int_{0}^{\pi/2} \ln (\sin x) d x \\ \end{array}\)

∴\( \mathop \smallint \limits_0^{\rm{\pi/2 }} {\rm{ln\;}}(\sin {\rm{x}}){\rm{\;dx}}=\rm{\frac{I}{2}} \)

Hence, option (4) is correct.