Theorems of Integral Calculus MCQ Quiz in தமிழ் - Objective Question with Answer for Theorems of Integral Calculus - இலவச PDF ஐப் பதிவிறக்கவும்

\(\mathop \smallint \limits_0^\pi {\sin ^3}\theta d\theta = I\)

\(\mathop \smallint \limits_0^\pi \left( {1 - {{\cos }^2}\theta } \right)\sin \theta d\theta\), let cosθ = t

⇒ -sinθ dθ = dt,

At θ = 0, t = cos 0 = 1

θ = π, t = cosπ = -1

\(\therefore I = - \mathop \smallint \limits_{ - 1}^{ + 1} \left( {1 - {t^2}} \right)dt = \left[ {t - \frac{{{t^3}}}{3}} \right]_{ - 1}^1\)

⇒ I = 4/3

∴ 3I = 4

Explanation:

\(\int_{\frac{cos x }{1-cosx^.}}^{}dx\) = \(\int_{\frac{cos x }{1-cosx}~\times~\frac{1+cosx}{1+cosx}}dx\)

= \(\int_{\frac{cos x (1+cosx) }{1-cos^2x}}dx~=~\int{\frac{cosx~+~cos^2x}{sin^2x}}dx\)
 \(=\int{(\frac{cosx}{sin^2x}~+~\frac{cos^2x}{sin^2x}})dx \\ =\int{(\frac{1}{sinx}\times \frac{cosx}{sinx}\times \frac{cos^2x}{sin^2x})}dx\)
 \(=\int{(𝑐𝑜𝑠𝑒𝑐 𝑥 . 𝑐𝑜𝑡𝑥 + cot^2 𝑥 )𝑑x}\\ =\int{(𝑐𝑜𝑠𝑒𝑐 𝑥 . 𝑐𝑜𝑡𝑥 + 𝑐𝑜𝑠𝑒𝑐^2𝑥 − 1 )𝑑x}\\ =\int{𝑐𝑜𝑠𝑒𝑐 𝑥 . 𝑐𝑜𝑡𝑥 }dx~+~\int{𝑐𝑜𝑠𝑒𝑐^2𝑥~ 𝑑x}~+~\int{1.dx}\)
\(\because~\int{𝑐𝑜𝑠𝑒𝑐^2𝑥 . 𝑑𝑥 = − cot 𝑥 + c}\\~and~\int{1 𝑑𝑥} = 𝑥 + c\)
∴ \(𝐼 = −𝑐𝑜𝑠𝑒𝑐 𝑥 − cot 𝑥 − 𝑥 + c\\ ~~=𝑐 − 𝑐𝑜𝑠𝑒𝑐 𝑥 − cot 𝑥 − x\)

Concept:

Integration Formulas:

\(\smallint {\rm{se}}{{\rm{c}}^2}{\rm{x\;dx}} = \tan {\rm{x}} + {\rm{C}}\)

\(\smallint \sec {\rm{x}}\tan {\rm{x\;dx}} = \sec {\rm{x}} + {\rm{c}}\)

Analysis:

Multiply and divide the given equation by (1 + sin x).

\(\smallint \frac{{{\rm{dx}}}}{{1 - \sin {\rm{x}}}} \times \frac{{1 + \sin {\rm{x}}}}{{1 + \sin {\rm{x}}}}{\rm{\;dx}} = {\rm{}}\smallint \frac{{1 + \sin {\rm{x}}}}{{1 - {\rm{si}}{{\rm{n}}^2}{\rm{x}}}}{\rm{dx}}\)

Since  1 - sin2x = cos2x, the above equation can be written as:

\( = {\rm{\;}}\smallint \frac{{1 + \sin {\rm{x}}}}{{{{\cos }^2}{\rm{x}}}}{\rm{dx}}\)

\( = \smallint \frac{1}{{{{\cos }^2}{\rm{x}}}}{\rm{dx}} + {\rm{}}\smallint \frac{{\sin {\rm{x}}}}{{{\rm{co}}{{\rm{s}}^2}{\rm{x}}}}{\rm{dx}}\)

\( = \smallint {\sec ^2}{\rm{xdx}} + {\rm{}}\smallint \sec {\rm{x}}\tan {\rm{xdx}}\)

We know that \(\smallint \frac{{{\rm{f'}}\left( {\rm{x}} \right)}}{{{\rm{f}}\left( {\rm{x}} \right)}}{\rm{\;dx}} = \log {\rm{f}}\left( {\rm{x}} \right) + {\rm{C}}\)

In the given problem \({\rm{f}}\left( {\rm{x}} \right) = {{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}\)

We know that \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}} \right) = \left( {{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}} \right)\)

f(x) = x²e^-x

f^I(x) = x²(-e^-x) + e^-x × 2x

⇒ f^I(x) = e^-x(2x-x²)

Putting f^I(x) = 0

⇒ e^-x (2x – x²) = 0

x(2-x) = 0

⇒ x = 0 or x = 2

∴ x = 0 & x = 2 are the stationary points

f^II(x) = e^-x(2 - 2x) + (2x – x²) (-e^-x)

= e^-x (x² – 4x + 2)

At x = 0, f^II(0) = 2 > 0

Explanation:

Then given family of curves is

\(y = cx^{-2k} \)---(i)

Let us first find the differential equation satisfied by the family (i).

For this, we differentiate (i). w.r.to x 

∴ \(\frac{{dy}}{{dx}} = -2ck{x^{-2k - 1}} \)

∴ The differential equation of the orthogonal trajectories will be obtained by replacing \(\frac{{dy}}{{dx}}\) by \(- \frac{{dx}}{{dy}}\).

⇒ Orthogonal trajectories are given by:

\(- \frac{{dx}}{{dy}} =-2 ck{x^{-2k - 1}} \)

⇒ \( \frac{{dx}}{{dy}} = 2\frac{{ck{x^{-2k}}}}{x} = 2\frac{{ky}}{x}\)

⇒ x dx - 2 k y dy = 0

⇒ \(\frac{1}{2}{x^2} - 2k \cdot \frac{1}{2}{y^2}\) = constant

⇒ x² - ky² = constant

Hence Option(4) is the correct answer.

(f'(x)) = 0

a/x + 2bx +1= 0

at  x=-1,  -a - 2b + 1 = 0

a + 2b = 1 ..........(1)

At x=2,  a/2 + 4b + 1 = 0

a + 8b = -2 ..........(2)

subtracting (1) & (2),   b=-1/2

=> a=2

The given function is odd function.

\(\begin{array}{l} f\left( x \right) = {x^3} + \frac{1}{x} + 1\\ f'\left( x \right) = 3{x^2} - \frac{1}{{{x^2}}}\\ f\left( 2 \right) = 3 \times {2^2} - \frac{1}{{{2^2}}} = 11.75 \end{array}\)