Theorems of Integral Calculus MCQ Quiz - Objective Question with Answer for Theorems of Integral Calculus - Download Free PDF

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.

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:

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

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

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

Given a differential equation,

\(\frac{dy}{dx}=e^{-3x}\)

or, dy = e^-3x dx

Integrated on both side,

∫dy = ∫e-3x dx

or, y = e-3x × \(\frac{-1}{3}\) + K

\(z=\sqrt{4-x^2}\)

\(\frac{\partial x}{\partial x}=\frac{-x}{\sqrt{4-x^2}},\frac{\partial z}{\partial y}=0\)

Required surface area = \(\displaystyle\int_{x=0}^1\int^4_{y=0}\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}dxdy\)

\(\displaystyle\int_{x=0}^1\int^4_{y=0}\sqrt{\left(\frac{-x}{\sqrt{4-x^2}}\right)^2+0^2+1}\ dxdy\)

\(\displaystyle\int_{x=0}^1\int^4_{y=0}\sqrt{\left(\frac{4}{\sqrt{4-x^2}}\right)}\ dxdy=2\displaystyle\int_{x=0}^1\frac{dx}{\sqrt{4-x^2}}\times\displaystyle\int_{y=0}^4(1)\ dy\)

\(=2\times\left[\sin^{-1}\left(\frac{x}{2}\right)\right]_0^1\times{4}\)

\(=8\times\left[\sin^{-1}\left(\frac{1}{2}\right)-\sin^{-1}(0)\right]\)

\(=8\left(\frac{\pi}{6}-0\right)=\frac{4}{3}.\pi=\frac{4\pi}{3}\)

Given:

\(\rm \displaystyle\int _0^{π/2} \sin x \ dx\)

The integration of sinx is -cosx

= [-cosx]₀^π/2

= 1

Additional InformationCalculus:

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.

Concept:

The product of function can be integrated by the method of "Integration by parts".

By the method of integration by parts we have,

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

Where f is the first function and g is the second function.

Which is chosen based on the order for the selection of the first function: ILATE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent)

Calculation:

Given:

We have to simplify the expression ∫ x log x dx

Here nothing has been mentioned so, we will take the base of the log as e.

And as we know, from integration by parts,

log x = u and x = v

which leads to,

\(I = \log x \times \frac{{{x^2}}}{2} - \frac{1}{2}\smallint {x^2} \times \frac{1}{x}dx\)

\(I = \log x \times \frac{{{x^2}}}{2} - \frac{1}{2} \times \frac{{{x^2}}}{2}\)

\(I=\frac{{{x^2}}}{2}\left( {\log x - \frac{1}{2}} \right)\)

Concept:

By using Greens theorem:

 \(\mathop \oint \limits_C^{} Mdx + Ndy\) = \(\;\mathop \int\!\!\!\int \limits_R^{} \left( {\frac{{∂ N}}{{∂ x}} - \frac{{∂ M}}{{∂ y}}} \right)dxdy\)

Calculation:

Given:

\(I = \mathop \oint \limits_C \left( {{x^2}y\;dy - {y^2}x\;dx} \right)\)

 where C is the boundary of square  0 ≤ x ≤ 1, 0 ≤ y ≤ 1

\(\mathop \oint \limits_C \left( {{x^2}y\;dy - {y^2}x\;dx} \right)\) = \(\;\mathop \int\!\!\!\int \limits_R^{} \left( {\frac{{∂ N}}{{∂ x}} - \frac{{∂ M}}{{∂ y}}} \right)dxdy\)

∂N/∂x = 2xy, ∂M/∂y = -2xy

using the values in the Greens theorem :

\(\;\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 \left( {2xy + 2xy} \right)dxdy\) = \( \smallint \limits_{x = 0}^1\smallint \limits_{y = 0}^1 \left( {4xy} \right)dxdy\)

\(\;\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 \left( {2xy + 2xy} \right)dxdy\) =\(\;4\mathop \smallint \limits_{x = 0}^1 x\left[ {\frac{{{y^2}}}{2}} \right]_0^1dx\)

=\(\;2\mathop \smallint \limits_{x = 0}^1 xdx\)

=\(\;2\left[ {\frac{{{x^2}}}{2}} \right]_0^1\)

= 1

Explanation:

\(let\ I = ∫_0^{(π /4)}x cos (x^2) dx\)

put t = x²

differentiatiing w.r.t to x

dt = 2xdx

\(\frac{dt}{2} = xdx\)

if x = 0, then t = 0

if x = \(\frac{π}{4}\) then  \(t=\frac{π^2}{16}\)

\( I = \frac{∫_0^{(π^2 /16)}cos (t)}{2} dt\)

\(I =0.5 [\sin t]_0^{\frac{π^2}{16}} \)

As mentioned in the question we should take π = 3.14

\(I = 0.5[\sin t]_0^{\frac{(3.14)^2}{16}} \)

\(I =0.5\left [\sin\left\{\frac{(3.14)^2}{16}\right\}-\sin 0 \right]\)

Please use the calculator in radian mode for trigonometric and exponential functions.

I = 0.2889 ≈ 0.289

Target is having inclination of 45° with x-axis

\(\begin{array}{l} \Rightarrow \frac{{dy}}{{dx}} = \tan 45^\circ \\ \Rightarrow \frac{{d\left( {xln\;x} \right)}}{{dx}} = 1\\ \Rightarrow \ln x + \frac{x}{x} = 1 \end{array}\)

\(z=\sqrt{4-x^2}\)

\(\frac{\partial x}{\partial x}=\frac{-x}{\sqrt{4-x^2}},\frac{\partial z}{\partial y}=0\)

Required surface area = \(\displaystyle\int_{x=0}^1\int^4_{y=0}\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}dxdy\)

\(\displaystyle\int_{x=0}^1\int^4_{y=0}\sqrt{\left(\frac{-x}{\sqrt{4-x^2}}\right)^2+0^2+1}\ dxdy\)

\(\displaystyle\int_{x=0}^1\int^4_{y=0}\sqrt{\left(\frac{4}{\sqrt{4-x^2}}\right)}\ dxdy=2\displaystyle\int_{x=0}^1\frac{dx}{\sqrt{4-x^2}}\times\displaystyle\int_{y=0}^4(1)\ dy\)

\(=2\times\left[\sin^{-1}\left(\frac{x}{2}\right)\right]_0^1\times{4}\)

\(=8\times\left[\sin^{-1}\left(\frac{1}{2}\right)-\sin^{-1}(0)\right]\)

\(=8\left(\frac{\pi}{6}-0\right)=\frac{4}{3}.\pi=\frac{4\pi}{3}\)

Given a differential equation,

\(\frac{dy}{dx}=e^{-3x}\)

or, dy = e^-3x dx

Integrated on both side,

∫dy = ∫e-3x dx

or, y = e-3x × \(\frac{-1}{3}\) + K

\(\frac{1}{{1 + tanx}} = \frac{1}{{1 + \frac{{sinx}}{{\cos x}}}} = \frac{{cosx}}{{sinx + cosx}}\)

we get,

\(\smallint \frac{{cosx\;dx}}{{sinx + cosx}} = \frac{1}{2}\smallint \left[ {\frac{{cosx + sinx + cosx - sinx}}{{cosx + sinx}}} \right]dx \)

\(\frac{1}{2}\smallint \left[ {1 + \frac{{cosx - sinx}}{{cosx + sinx}}} \right]dx\)

\( = \frac{1}{2}\left[ {x + \;log\;\left| {sinx + cosx} \right|} \right] + C\)

\(\mathop \smallint \limits_{1/\pi }^{2/\pi } \frac{{\cos \left( {1/x} \right)}}{{{x^2}}}dx\)

\(\mathop \smallint \limits_{1/\pi }^{2/\pi } \frac{{\cos \left( {1/x} \right)}}{{{x^2}}}dx\; = \;\mathop \smallint \limits_{2/\pi }^{1/\pi } \frac{{ - \cos \left( {1/x} \right)}}{{{x^2}}}dx\)

\(= \;\mathop \smallint \limits_{\frac{\pi }{2}}^\pi {\rm{cosy}}dy\; = \;\left| {siny} \right|_{\pi /2}^\pi\)

= 0 - 1 = - 1

Concept:

The product of function can be integrated by the method of "Integration by parts".

By the method of integration by parts we have,

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

Where f is the first function and g is the second function.

Which is chosen based on the order for the selection of the first function: ILATE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent)

Calculation:

From the above formula,

\(\int x^2e^{ax}dx=\frac{x^2e^{ax}}{a}-\int \frac{2xe^{ax}}{a}dx\)

⇒ \(x^2e^{ax}-\frac{2}{a}[\frac{xe^{ax}}{a}-\frac{e^{ax}}{a^2}]\)

⇒ \(x^2e^{ax}-\frac{2}{a^2}xe^{ax}+\frac{2}{a^3}e^{ax}\)

Hence, \(\rm \displaystyle\int x^2 e^{ax} dx\) = \(\rm \frac{e^{ax}}{a^3} (a^2 x^2 -2ax + 2)\)