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

$y=c{x}^{-2k}$---(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:

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

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

⇒ 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:

${\int }_{\frac{cosx}{1-cos{x}^{.}}}^{}dx$ = ${\int }_{\frac{cosx}{1-cosx}\times \frac{1+cosx}{1+cosx}}dx$

= ${\int }_{\frac{cosx(1+cosx)}{1-co{s}^{2}x}}dx=\int \frac{cosx+co{s}^{2}x}{si{n}^{2}x}dx$
 $$\begin{gathered} =\int (\frac{cosx}{si{n}^{2}x}+\frac{co{s}^{2}x}{si{n}^{2}x})dx \\ =\int (\frac{1}{sinx}\times \frac{cosx}{sinx}\times \frac{co{s}^{2}x}{si{n}^{2}x})dx \end{gathered}$$
 $$\begin{gathered} =\int (𝑐𝑜𝑠𝑒𝑐𝑥.𝑐𝑜𝑡𝑥+co{t}^2𝑥)𝑑x \\ =\int (𝑐𝑜𝑠𝑒𝑐𝑥.𝑐𝑜𝑡𝑥+𝑐𝑜𝑠𝑒{𝑐}^2𝑥-1)𝑑x \\ =\int 𝑐𝑜𝑠𝑒𝑐𝑥.𝑐𝑜𝑡𝑥dx+\int 𝑐𝑜𝑠𝑒{𝑐}^2𝑥𝑑x+\int 1.dx \end{gathered}$$
$$\begin{gathered} \because \int 𝑐𝑜𝑠𝑒{𝑐}^2𝑥.𝑑𝑥=-cot𝑥+c \\ and\int 1𝑑𝑥=𝑥+c \end{gathered}$$
∴ $$\begin{gathered} 𝐼=-𝑐𝑜𝑠𝑒𝑐𝑥-cot𝑥-𝑥+c \\ =𝑐-𝑐𝑜𝑠𝑒𝑐𝑥-cot𝑥-x \end{gathered}$$

Concept:

Integration Formulas:

$\int \mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x}\mathrm{d}\mathrm{x}=\tan \mathrm{x}+\mathrm{C}$

$\int \sec \mathrm{x}\tan \mathrm{x}\mathrm{d}\mathrm{x}=\sec \mathrm{x}+\mathrm{c}$

Analysis:

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

$\int \frac{\mathrm{d}\mathrm{x}}{1-\sin \mathrm{x}}\times \frac{1+\sin \mathrm{x}}{1+\sin \mathrm{x}}\mathrm{d}\mathrm{x}=\int \frac{1+\sin \mathrm{x}}{1-\mathrm{s}\mathrm{i}{\mathrm{n}}^2\mathrm{x}}\mathrm{d}\mathrm{x}$

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

$=\int \frac{1+\sin \mathrm{x}}{{\cos }^2\mathrm{x}}\mathrm{d}\mathrm{x}$

$=\int \frac{1}{{\cos }^2\mathrm{x}}\mathrm{d}\mathrm{x}+\int \frac{\sin \mathrm{x}}{\mathrm{c}\mathrm{o}{\mathrm{s}}^2\mathrm{x}}\mathrm{d}\mathrm{x}$

$=\int {\sec }^2\mathrm{x}\mathrm{d}\mathrm{x}+\int \sec \mathrm{x}\tan \mathrm{x}\mathrm{d}\mathrm{x}$

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

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

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

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

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

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

Given:

${\displaystyle {\int }_0^{\pi /2}\sin \mathrm{x}\mathrm{d}\mathrm{x}}$

The integration of sinx is -cosx

= [-cosx]₀^π/2

= 1

Additional InformationCalculus:

Derivatives of Trigonometric Functions:

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\sin \mathrm{x}=\cos \mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\cos \mathrm{x}=-\sin \mathrm{x}$

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\tan \mathrm{x}={\sec }^2\mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\cot \mathrm{x}=-{\csc }^2\mathrm{x}$

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\sec \mathrm{x}=\tan \mathrm{x}\sec \mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\csc \mathrm{x}=-\cot \mathrm{x}\csc \mathrm{x}$

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,

$\int f\left(x\right)g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx-\int \left[f^{\prime }\left(x\right)\int g\left(x\right)dx\right]dx$

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,

$I=\log x\times \frac{x^2}{2}-\frac{1}{2}\int x^2\times \frac{1}{x}dx$

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

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

Concept:

By using Greens theorem:

 $\underset{C}{\overset{}{\oint }}Mdx+Ndy$ = $\int \underset{R}{\overset{}{\int }}\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dxdy$

Calculation:

Given:

$I=\underset{C}{\oint }\left(x^{2}ydy-y^{2}xdx\right)$

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

$\underset{C}{\oint }\left(x^{2}ydy-y^{2}xdx\right)$ = $\int \underset{R}{\overset{}{\int }}\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dxdy$

∂N/∂x = 2xy, ∂M/∂y = -2xy

using the values in the Greens theorem :

$\underset{x=0}{\overset{1}{\int }}\underset{y=0}{\overset{1}{\int }}\left(2xy+2xy\right)dxdy$ = $\underset{x=0}{\overset{1}{\int }}\underset{y=0}{\overset{1}{\int }}\left(4xy\right)dxdy$

$\underset{x=0}{\overset{1}{\int }}\underset{y=0}{\overset{1}{\int }}\left(2xy+2xy\right)dxdy$ =$4\underset{x=0}{\overset{1}{\int }}x{\left[\frac{y^2}{2}\right]}_0^{1}dx$

=$2\underset{x=0}{\overset{1}{\int }}xdx$

=$2{\left[\frac{x^2}{2}\right]}_0^1$

= 1

Explanation:

$letI={\int }_0^{(\pi /4)}xcos(x^2)dx$

put t = x²

differentiatiing w.r.t to x

dt = 2xdx

$\frac{dt}{2}=xdx$

if x = 0, then t = 0

if x = $\frac{\pi }{4}$ then  $t=\frac{{\pi }^2}{16}$

$I=\frac{{\int }_0^{({\pi }^2/16)}cos(t)}{2}dt$

$I=0.5[\sin t{]}_0^{\frac{{\pi }^2}{16}}$

As mentioned in the question we should take π = 3.14

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

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

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

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

$z=\sqrt{4-x^2}$

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

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

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

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

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

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

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

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,

$\int \frac{cosxdx}{sinx+cosx}=\frac{1}{2}\int \left[\frac{cosx+sinx+cosx-sinx}{cosx+sinx}\right]dx$

$\frac{1}{2}\int \left[1+\frac{cosx-sinx}{cosx+sinx}\right]dx$

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

$\underset{1/\pi }{\overset{2/\pi }{\int }}\frac{\cos \left(1/x\right)}{x^2}dx$

$\underset{1/\pi }{\overset{2/\pi }{\int }}\frac{\cos \left(1/x\right)}{x^2}dx=\underset{2/\pi }{\overset{1/\pi }{\int }}\frac{-\cos \left(1/x\right)}{x^2}dx$

$=\underset{\frac{\pi }{2}}{\overset{\pi }{\int }}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{y}dy={\left|siny\right|}_{\pi /2}^{\pi }$

= 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,

$\int f\left(x\right)g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx-\int \left[f^{\prime }\left(x\right)\int g\left(x\right)dx\right]dx$

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,

$\int x^2{e}^{ax}dx=\frac{x^2{e}^{ax}}{a}-\int \frac{2x{e}^{ax}}{a}dx$

⇒ $x^2{e}^{ax}-\frac{2}{a}[\frac{x{e}^{ax}}{a}-\frac{e^{ax}}{a^2}]$

⇒ $x^2{e}^{ax}-\frac{2}{a^2}x{e}^{ax}+\frac{2}{a^3}{e}^{ax}$

Hence, ${\displaystyle \int {\mathrm{x}}^2{\mathrm{e}}^{\mathrm{a}\mathrm{x}}\mathrm{d}\mathrm{x}}$ = $\frac{{\mathrm{e}}^{\mathrm{a}\mathrm{x}}}{{\mathrm{a}}^3}({\mathrm{a}}^2{\mathrm{x}}^2-2\mathrm{a}\mathrm{x}+2)$