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

$\underset{0}{\overset{\pi }{\int }}{\sin }^3\theta d\theta =I$

$\underset{0}{\overset{\pi }{\int }}\left(1-{\cos }^2\theta \right)\sin \theta d\theta$, let cosθ = t

⇒ -sinθ dθ = dt,

At θ = 0, t = cos 0 = 1

θ = π, t = cosπ = -1

$\therefore I=-\underset{-1}{\overset{+1}{\int }}\left(1-t^2\right)dt={\left[t-\frac{t^3}{3}\right]}_{-1}^1$

⇒ I = 4/3

∴ 3I = 4

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:

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

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

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

We know that $\int \frac{{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}\mathrm{d}\mathrm{x}=\log \mathrm{f}\left(\mathrm{x}\right)+\mathrm{C}$

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

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

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

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

(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.

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