Mathematics Test - 18

Let,

 \(I = \smallint {e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx\)

\( = \smallint {e^x}f\left( x \right)dx + \smallint {e^x}f'\left( x \right)dx+C\)

By solving through integration by parts, we get

\( = \left\{ {{e^x}f\left( x \right) - \smallint f'\left( x \right){e^x}dx} \right\} + \smallint {e^x}f'\left( x \right)dx +C\)

\( = f\left( x \right).{e^x} +C\)

where C is constant

Formula used:

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

tan(π - θ) = - tan θ  

Calculation:

Let

I = \(\int^π _0 ln\left(tan\frac{x}{2}\right)dx\)        ---(1)

According to the formula used

I = \(\int^π _0 ln\left(tan (π -\frac{x}{2})\right)dx\)

⇒ I = - \(\int^π _0 ln\left(tan\frac{x}{2}\right)dx\)

From equation (1)

⇒ I = -I

⇒ 2I = 0

⇒ I = 0

∴ The value of the integral \(\int^π _0 ln\left(tan\frac{x}{2}\right)dx\) is 0.

Concept:

If f(x)  is odd function then f(-x) = -f(x)

Properties of definite integral

If f(x)  is even function then \(\mathop \smallint \nolimits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = 2\mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}}\)

If f(x)  is odd function then \(\mathop \smallint \nolimits_{ - {\rm{a}}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} =0\)

Calculation:

Let I = \(\displaystyle\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \rm \dfrac{sin^5 \ x \ cos^3 \ x}{x^4}dx\)

Let f(x) = \(\rm \dfrac{sin^5 \ x \ cos^3 \ x}{x^4}\)

Replaced x by -x, 

⇒ f(-x) = \(\rm \dfrac{sin^5 \ (-x) \ cos^3 \ (-x)}{(-x)^4}\)

As we know sin (-θ) = - sin θ and cos (-θ) = cos θ

= \(\rm \dfrac{-sin^5 \ x \ cos^3 \ x}{x^4}\)

⇒ f(-x) = -f(x)      

So, f(x) is odd function

Therefore, I = 0     

CONCEPT:

• The limit of a sum given in the general form can be converted into a definite integral form by replacing r/n with x and 1/n with dx.
• We can see that here n is very large and tends to ∞ so the reciprocal of n will be very small and can be replaced as dx.

\(I = \rm \lim_{n \rightarrow ∞} ∑_{g(n) }^{h(n)} \frac{f ( r/n)}{n} = \int_a^b f(x) dx\)

• For the upper and lower limits of the definite integral,

\(\rm a = \lim_{n \rightarrow ∞} \frac{g(n)}{n}\)

\(b = \ln_{n \rightarrow ∞} \frac{h(n)}{n}\)

∑ is Replaced by the sign of Integration

\(\frac{r}{n}\) is Replaced by x

\(\frac{1}{n}\) is replaced by dx

CALCULATION:

Given:

\(I = \lim_{n \rightarrow \infty} \sum_{r = 0}^{r = n} \frac{1}{n} sin \left( \frac{r}{n} \right) dx\)

• For the upper and lower limits of the definite integral,

\(\rm a = \lim_{n \rightarrow ∞} \frac{n}{n} = 1\)

\(b = \ln_{n \rightarrow ∞} \frac{0}{n} = 0\)

\(⇒ I = \int_0^1 \sin x dx\)

⇒ I = \((-\cos x)_0^1\) = - (cos 1 - (cos 0))

⇒ I = 1 - cos 1

So, the correct answer is option 1

Concept:

Definite Integral:

If ∫ f(x) dx = g(x) + C, then \(\rm \int_a^b f(x)\ dx = \left[ g(x)\right]_a^b\) = g(b) - g(a).

Properties:

• For even functions: f(-x) = f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).
• For odd functions: f(-x) = -f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=0\).
• \(\rm \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx\).
• If f(x) = f(2a - x), then \(\rm \int_0^{2a}f(x)\ dx=2\int_0^af(x)\ dx\).

Calculation:

It can be observed that sin (-θ) = -sin θ.

Let f(x) = |sin x|

Put x = -x

⇒ f(-x) = |sin -x| = |-sin x| = |sin x| = f(x)

∴ |sin x| is an even function.

Therefore, using the properties of definite integrals, we get:

\(\rm \int_{-\pi/2}^{\pi/2}|\sin x|\ dx\)

= 2\(\rm \int_{0}^{\pi/2}\sin x\ dx\)

= 2\(\rm \left[-\cos x\right]_{0}^{\pi/2}\)

= -2\(\rm \left[\cos \pi/2-\cos 0\right]\)

= -2[0 - 1]

= 2.

CONCEPT:

Fundamental Theorem of Calculus (Newton-Leibniz formula)

\(\rm \frac{d}{dx} \left[ \int_{g(x)}^{h(x)} f(t) dt \right] = f(h(x)) . h'(x) - f(g(x)) g'(x)\)

CONCEPT:

Given:​

\(I = \rm \int_{\sin x}^{\cos x} \log t^2 \ dt\)

Using the Newton-Lebenitz formula

\(\rm \frac{d}{dx} \left[ \int_{\sin x}^{\cos x} \log t^2 dt \right] = \frac{dI}{dt}\)

\(⇒ \rm \frac{d(I)}{dx} = [\log (\cos x)^2 ] \frac{d}{dx} \cos x - [ \log (\sin x)^2 \frac{d}{dx} \sin x\)

\(⇒ \rm \frac{d(I)}{dx} \rm = [log \cos^2 x] (- \sin x) - [\log sin^2 x] \cos x\)

⇒ dI/dt= -2 sin x log cos x - 2 cos x log sin x

\(\Rightarrow \frac{dI}{dx} (x = π /4) = -2 \sin\frac{π}{4} \log \cos \frac{π}{4} - 2 \cos \frac{π}{4} \log \sin \frac{π}{4}\)

\(\Rightarrow \frac{dI}{dx} (x = π /4)= - \frac{2}{√ 2} \log \frac{1}{√ 2} - \frac{2}{√ 2} \log \frac{1}{√ 2}\)

\(\Rightarrow \frac{dI}{dx} (x = π /4)= - √ 2 \left( \log \left( \frac{1}{√ 2} \times \frac{1}{√ 2} \right) \right) \) (∵  log m + log n = log mn)

⇒ dI/dt at π/4 = -√2 log \(\frac{1}{2}\)

⇒ dI/dt at π/4 = √2 log 2

So, the correct answer is option 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:

∫cosx dx = sinx + c

∫ secx dx = ln(sec x + tan x) + c

cos2θ = 2cos²θ - 1

\(\rm 1\over cos θ \)= secθ 

Calculation:

Given integration is, \(\rm ∫{ {cos2x}\over {cosx} }dx\)

= \(\rm ∫{ {2cos^2x-1}\over {cosx} }dx\)

= \(\rm ∫({ {{2cos^2x}\over {cosx}}-{{1}\over {cosx} }})dx\)

= \(\rm ∫({ {{2cosx}}-{secx} })dx\)

= 2sinx - ln(sec x + tan x) + C, C = constant of integration

Formula used:

\(\int_{a}^{b}f(x)dx\ =\ \int_{a}^{b}f(a\ +\ b\ -\ x)dx\)

\(sin(\frac{\pi}{2}\ -\ \theta)\ =\ cos\theta\)

\(cos(\frac{\pi}{2}\ -\ \theta)\ =\ sin\theta\)

Calculation:

Given that,

\(\int_{0}^{\frac{\pi}{2}}\frac{sinx}{sinx\ +\ cosx}dx\ =\ g''(x)\ +\ C\)         ----(1)

Using the property \(\int_{a}^{b}f(x)dx\ =\ \int_{a}^{b}f(a\ +\ b\ -\ x)dx\)

\(\int_{0}^{\frac{\pi}{2}}\frac{sin(\frac{\pi}{2}\ -\ x)}{sin(\frac{\pi}{2}\ -\ x)\ +\ cos(\frac{\pi}{2}\ -\ x)}dx\ =\ g''(x)\ +\ C\)

\(\Rightarrow \ \int_{0}^{\frac{\pi}{2}}\frac{cosx}{sinx\ +\ cosx}dx\ =\ g''(x)\ +\ C\)        ----(1)

Adding both equation, we will get

\(\Rightarrow \ \int_{0}^{\frac{\pi}{2}}(\frac{sinx}{sinx\ +\ cosx}\ +\ \frac{cosx}{sinx\ +\ cosx})dx=\ 2[g''(x)\ +\ C]\)

\(\Rightarrow \ \int_{0}^{\frac{\pi}{2}}\frac{sinx\ +\ cosx}{sinx\ +\ cosx}dx\ =\ 2[g''(x)\ +\ C]\)

\(\Rightarrow \ \int_{0}^{\frac{\pi}{2}}dx\ =\ 2[g''(x)\ +\ C]\)

\(\Rightarrow \ (x)_0^\frac{\pi}{2} =\ 2[g''(x)\ +\ C]\)

\(\Rightarrow \ \frac{\pi}{4} =\ g''(x)\ +\ C\)

Integrating both side with respect to x

\(\Rightarrow \ \frac{\pi}{4}x =\ g'(x)\ +\ Cx\)

Therefore, put x = 1 on above equation 

\(\Rightarrow \ \frac{\pi}{4} =\ g'(1)\ +\ C\)        ----(3)

But according to question,  g'(1) = 1

\(\Rightarrow \ C\ =\ \frac{\pi}{4} \ -\ 1\)

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