Integral Calculus MCQ Quiz in मल्याळम - Objective Question with Answer for Integral Calculus - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Calculation:

Given, Curves are y² = 2y - x and y-axis.

⇒ x = 2y - y² ...(1)     and x = 0 ....(2)

From (1) and (2), we get,

2y - y² = 0

⇒ y(2 - y) = 0

⇒ y = 0 or y = 2

Thus the curve meets the y-axis at (0,0) and (0,2)

So, the required area = ${\int }_0^{2}x.dy$

$={\int }_0^2(2y-y^2)dy$

$={\int }_0^{2}2ydy-{\int }_0^2{y}^{2}dy$

$=[y^2{]}_0^2-{\left[\frac{y^3}{3}\right]}_0^2$

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

= $\frac{4}{3}$

Concept:
$\int \frac{1}{a}.da=loga$      ----(1)

Calculation:

$\int \frac{x^{3}dx}{x+1}$

$\Rightarrow \int \frac{x^3+1-1}{x+1}.dx$

$\Rightarrow \int \frac{x^3+1}{x+1}.dx-\int \frac{1}{x+1}.dx$

$\Rightarrow \int \frac{(x+1)(x^2+1-x)}{x+1}.dx-\int \frac{1}{x+1}.dx$

⇒ ∫ (x² + 1 - x)dx - log (x + 1)      [using (1)]

$\Rightarrow x-\frac{x^2}{2}+\frac{x^3}{3}-\log |1+x|+c$

Concept:

f(x) = |x| will be equal to 

• x, if x > 0
• -x, if x < 0
• 0, if x = 0

∫ dx = x + C  (C is a constant)

∫ xⁿ dx = xⁿ⁺¹/n+1 + C

Calculation:

Let, $I={\int }_{-2}^{-1}\frac{x}{|x|}dx$

Using the above concept, as x ∈ (-2, -1)

⇒ $I={\int }_{-2}^{-1}\frac{x}{-x}dx$

⇒ $I=-1{\int }_{-2}^{-1}(1)dx$

⇒ $I=-[x{]}_{-2}^{-1}$  

⇒ I = -[-1 - (-2)] 

∴  ${\int }_{-2}^{-1}\frac{x}{|x|}dx$ = -1

Concept:

• cos 2x = cos² x - sin² x

Calculation:

$\int \frac{\cos 2x}{{\cos }^{2}x.{\sin }^{2}x}dx$

= $\int \frac{co{s}^{2}x-si{n}^{2}x}{{\cos }^{2}x.{\sin }^{2}x}dx$

= $\int (\frac{1}{si{n}^{2}x}-\frac{1}{co{s}^{2}x})dx$

= $\int \frac{1}{{\sin }^{2}x}dx-\int \frac{1}{{\cos }^{2}x}dx$

= $\int \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x}\mathrm{d}\mathrm{x}-\int \mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x}\mathrm{d}\mathrm{x}$

= - cot x - tan x + C

Concept:

${\displaystyle \int {\displaystyle \frac{1}{\mathrm{x}}}\mathrm{d}\mathrm{x}=\log \mathrm{x}+\mathrm{c}}$

Calculation:

$$\begin{gathered} \text{Let I}={\displaystyle \int {\displaystyle \frac{1}{1+{\mathrm{e}}^{\mathrm{x}}}}\mathrm{d}\mathrm{x}} \\ ={\displaystyle \int {\displaystyle \frac{{\mathrm{e}}^{-\mathrm{x}}}{{\mathrm{e}}^{-\mathrm{x}}+1}}\mathrm{d}\mathrm{x}} \end{gathered}$$

Assume e^-x + 1 = t

Differenatiang with respect to x, we get

⇒ -e^-x dx = dt

∴ e-x dx = -dt

$$\begin{gathered} ={\displaystyle \int {\displaystyle \frac{-\mathrm{d}\mathrm{t}}{\mathrm{t}}}} \\ =-\log \mathrm{t}+\mathrm{c} \\ =-\log ({\mathrm{e}}^{-\mathrm{x}}+1)+\mathrm{c} \\ =\log \left(\frac{1}{{\mathrm{e}}^{-\mathrm{x}}+1}\right)+\mathrm{c} \\ =\log \left(\frac{{\mathrm{e}}^{\mathrm{x}}}{1+{\mathrm{e}}^{\mathrm{x}}}\right)+\mathrm{c} \end{gathered}$$

Calculation:

Enclosed Area

$=2{\int }_0^{\pi /4}\left(\cos x-\sin x\right)dx$

$=2{\left[\sin x+\cos x\right]}_0^{\pi /4}$

$=2\left[\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-\left(0+1\right)\right]$

$=2\left(\sqrt{2}-1\right)$

Concept:

$\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$

Calculation:

I = $\int \sqrt{4\mathrm{x}-3}\mathrm{d}\mathrm{x}$

Let 4x - 3 = t²

Differenating with respect to x, we get

⇒ 4dx = 2tdt

⇒ dx = $\frac{\mathrm{t}}{2}$dt

Now,

I = $\int \sqrt{{\mathrm{t}}^2}\times \frac{\mathrm{t}}{2}\mathrm{d}\mathrm{t}$

= $\frac{1}{2}\int {\mathrm{t}}^2\mathrm{d}\mathrm{t}$

= $\frac{{\mathrm{t}}^3}{6}+\mathrm{c}$

= $\frac{(4\mathrm{x}-3{)}^{3/2}}{6}+\mathrm{c}$

Concept:

$\int \frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}=\log \mathrm{x}+\mathrm{c}$

Calculation:

I = $\int \frac{\mathrm{x}}{3{\mathrm{x}}^2+4}\mathrm{d}\mathrm{x}$

Let 3x² + 4 = t

Differentiating with respect to x, we get

⇒ 6xdx = dt

⇒ xdx = $\frac{\mathrm{d}\mathrm{t}}{6}$

Now,

I = $\frac{1}{6}\int \frac{1}{\mathrm{t}}\mathrm{d}\mathrm{t}$

= $\frac{1}{6}\log \mathrm{t}+\mathrm{c}$

= $\frac{1}{6}\log (3{\mathrm{x}}^2+4)+\mathrm{c}$

Concept:

• cos2x = 1 - 2sin²x ⇒ $si{n}^{2}x=\frac{1-cos2x}{2}$
• sin(-x) = -sinx

Calculation:

$={\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}\frac{1-cos2x}{2}dx$

$=\frac{1}{2}{\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}(1-cos2x)dx$

$=\frac{1}{2}[{\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}1.dx-{\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}cos2xdx]$

$=\frac{1}{2}[(x{)}_{\frac{-\pi }{3}}^{\frac{\pi }{3}}-{\left(\frac{sin2x}{2}\right)}_{\frac{-\pi }{3}}^{\frac{\pi }{3}}]$

$=\frac{1}{2}[\frac{2\pi }{3}-\frac{1}{2}(sin\frac{2\pi }{3}-sin\frac{-2\pi }{3})]$

$=\frac{1}{2}[\frac{2\pi }{3}-\frac{1}{2}(sin\frac{2\pi }{3}+sin\frac{2\pi }{3})]$

$=\frac{1}{2}[\frac{2\pi }{3}-\frac{1}{2}\left(2sin\frac{2\pi }{3}\right)]$

$=\frac{1}{2}[\frac{2\pi }{3}-sin\frac{2\pi }{3}]$

$=\frac{1}{2}[\frac{2\pi }{3}-\frac{\sqrt{3}}{2}]$

$=\frac{\pi }{3}-\frac{\sqrt{3}}{4}$

Concept:

         $\int u.vdx=u.\int vdx-\int (u^{\prime }\int vdx)dx$

Calculation:

Given, I =  $\int \frac{\log x}{{(x+1)}^2}dx$

If u(x) = log x and v(x) = $\frac{1}{(x+1{)}^2}$

⇒ I = $\log x\int \frac{1}{(x+1{)}^2}dx-\int (\frac{1}{x}\int \frac{1}{(x+1{)}^2}dx)dx$

⇒ I = $-\frac{\log x}{(x+1)}-\int \frac{-1}{x(x+1)}dx$

⇒ I = $-\frac{\log x}{(x+1)}+\int (\frac{1}{x}-\frac{1}{x+1})dx$

⇒ I = $-\frac{\log x}{(x+1)}+\log x-\log (x+1)+c$

∴ The correct answer is option (1).