3 Marks Question

Question 513 Marks

Integrate the function: $\frac{x}{\sqrt{x+4}}, x>0$

Answer

Let x + 4 = t
$\Rightarrow$ dx = dt
$\Rightarrow \int \frac{x}{\sqrt{x+4}} d x=\int \frac{(t-4)}{\sqrt{t}} d t$
= $\int\left(\sqrt{t}-\frac{4}{\sqrt{t}}\right) d t$
= $\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)+C$
= $\frac{2}{3}(t)^{\frac{3}{2}}-8(t)^{\frac{1}{2}}+C$
= $ \frac{2}{3} t \cdot t^{\frac{1}{2}}-8(t)^{\frac{1}{2}}+C$
= $\frac{2}{3}(x+4)^{\frac{1}{2}}(x+4-12)+C$
= $ \frac{2}{3} \sqrt{x+4}(x-8)+C$.

View full question & answer→

Question 523 Marks

Find an anti derivative (or integral) of the function by the method of inspection sin $2x - 4e^{3x}$

Answer

We know that, $\frac{d}{{dx}}\left( {\cos 2x} \right) = - 2\sin 2x$
$ \Rightarrow \frac{1}{{ - 2}}\frac{d}{{dx}}\left( {\cos 2x} \right) = \sin 2x$
$\Rightarrow \frac{d}{{dx}}\left( {\frac{{ - 1}}{2}\cos 2x} \right) = \sin 2x$ ...(i)
Again $\frac{d}{{dx}}{e^{3x}} = 3{e^{3x}}$
$\Rightarrow \frac{d}{{dx}}\left( {\frac{1}{3}{e^{3x}}} \right) = {e^{3x}}$
$\frac{d}{{dx}}\left( {\frac{{ - 4}}{3}{e^{3x}}} \right) = - 4{e^{3x}}$ [Multiplying both sides by -4] ...(ii)
Adding eq. (i) and (ii), we get
$ \Rightarrow \frac{d}{{dx}}\left( {\frac{{ - 1}}{2}\cos 2x} \right) + \frac{d}{{dx}}\left( {\frac{{ - 4}}{3}{e^{3x}}} \right) = sin\ 2x + (-4e^{3x})$
$ \Rightarrow \frac{d}{{dx}}\left( {\frac{{ - 1}}{2}\cos 2x - \frac{4}{3}{e^{3x}}} \right) = sin\ 2x - 4e^{3x}$
$\therefore $ An anti-derivative of sin $2x - 4e^{3x}$ is $\frac{{ - 1}}{2}\cos 2x - \frac{4}{3}{e^{3x}}.$

View full question & answer→

Question 533 Marks

Find the integral: $\int \sqrt{x}\left(3 x^{2}+2 x+3\right) d x$

Answer

$\int \sqrt{x}\left(3 x^{2}+2 x+3\right) d x$
= $\int\left(3 x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+3 x^{\frac{1}{2}}\right) d x$
= $3\left(\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\right)+2\left(\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right)+3\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$
= $\frac{6}{7} x^{\frac{7}{2}}+\frac{4}{5} x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+C$

View full question & answer→

Question 543 Marks

Find the integral: $\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x$

Answer

Let I =$\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x$
Separating the terms we get,
I =$\int\left(x^{\frac{5}{2}}+3 x^{\frac{1}{2}}+4 x^{-\frac{1}{2}}\right) d x$
Applying the formula,
$\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$
I = $\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+\frac{3\left(x^{\frac{3}{2}}\right)}{\frac{3}{2}}+\frac{4\left(x^{\frac{1}{2}}\right)}{\frac{1}{2}}+C$
= $\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 x^{\frac{1}{2}}+C$
= $\frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 \sqrt{x}+C$

View full question & answer→

Question 553 Marks

By using the properties of definite integrals, evaluate the integral $\int\limits_0^2 {x\sqrt {2 - x} dx} $

Answer

Let $I = \int\limits_0^2 {x\sqrt {2 - x} dx} $

$= \int\limits_0^2 {\left( {2 - x} \right)\sqrt {2 - \left( {2 - x} \right)} dx} $

$\left[ {\because \int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)dx } } } \right]$

$\Rightarrow I = \int\limits_0^2 {\left( {2 - x} \right)\sqrt x dx} $

$= \int\limits_0^2 {\left( {2{x^{\frac{1}{2}}} - {x^{\frac{3}{2}}}} \right)dx} $

$= \left[ {2.\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}} - \frac{{{x^{\frac{5}{2}}}}}{{\frac{5}{2}}}} \right]_0^2$

$= \left( {\frac{4}{3}{{.2}^{\frac{3}{2}}} - \frac{2}{5}{{.2}^{\frac{5}{2}}}} \right) - \left( {0 - 0} \right)$

$\Rightarrow I = \frac{4}{3} \times 2\sqrt 2 - \frac{2}{5} \times 4\sqrt 2 $

$ = \left( {\frac{8}{3} - \frac{8}{5}} \right)\sqrt 2 $

$= \frac{{16\sqrt 2 }}{{15}}$

View full question & answer→

Question 563 Marks

By using the properties of definite integrals, evaluate the integral $\int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} $

Answer

Let $I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} $
$= \int\limits_0^1 {\left( {1 - x} \right)\left\{ {1 - {{\left( {1 - x} \right)}}} \right\}^ndx} $
$\left[ {\because \int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)dx } } } \right]$
$\Rightarrow I = \int\limits_0^1 {\left( {1 - x} \right){{\left( {1 - 1 + x} \right)}^n}dx} $
$ = \int\limits_0^1 {\left( {1 - x} \right){x^n}dx} $
$= \int\limits_0^1 {\left( {{x^n} - {x^{n + 1}}} \right)dx} $
$\Rightarrow I = \left( {\frac{{{x^{n + 1}}}}{{n + 1}} - \frac{{{x^{n + 2}}}}{{n + 2}}} \right)_0^1$
$ = \frac{1}{{n + 1}} - \frac{1}{{n + 2}} - \left( {0 - 0} \right)$
$= \frac{{n + 2 - n - 1}}{{\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}$

View full question & answer→

Question 573 Marks

By using the properties of definite integrals, evaluate the integral $\int\limits_2^8 {\left| {x - 5} \right|dx} $

Answer

Let $I = \int\limits_2^8 {\left| {x - 5} \right|dx} $…(i)
Putting x - 5 = 0
$ \Rightarrow x = 5 \in \left( {2,8} \right)$
$\therefore$ From eq. (i),
$I = \int\limits_2^5 {\left| {x - 5} \right|dx + \int\limits_5^8 {\left| {x + 5} \right|dx} } $
$= \int\limits_2^5 { - \left( {x - 5} \right)dx + \int\limits_5^8 {\left( {x - 5} \right)dx} } $
$= - \left( {\frac{{{x^2}}}{2} - 5x} \right)_2^5 + \left( {\frac{{{x^2}}}{2} - 5x} \right)_{^5}^8$
$= - \left[ {\left( {\frac{{25}}{2} - 25} \right) - \left( {2 - 10} \right)} \right] + \left[ {\left( {32 - 40} \right) - \left( {\frac{{25}}{2} - 25} \right)} \right]$
$= - \left( { - \frac{{25}}{2} + 8} \right) + \left( { - 8 + \frac{{25}}{2}} \right)$
$= \frac{{25}}{2} - 8 - 8 + \frac{{25}}{2}$
= 25 - 16
= 9

View full question & answer→

Question 583 Marks

By using the properties of definite integrals, evaluate the integral $\int\limits_{ - 5}^5 {\left| {x + 2} \right|dx} $

Answer

Let $I = \int\limits_{ - 5}^5 {\left| {x + 2} \right|dx} $ …(i)

Putting x + 2 = 0

$ \Rightarrow x = - 2 \in \left( { - 5,5} \right)$

$\therefore$ From eq. (i),

$I = \int\limits_{ - 5}^{ - 2} {\left| {x + 2} \right|} dx + \int\limits_{ - 2}^5 {\left| {x + 2} \right|dx} $

$= \int\limits_{ - 5}^{ - 2} { - \left( {x + 2} \right)dx + \int\limits_{ - 2}^5 {\left( {x + 2} \right)dx} } $

$= - \left( {\frac{{{x^2}}}{2} + 2x} \right)_{ - 5}^{ - 2} + \left( {\frac{{{x^2}}}{2} + 2x} \right)_{ - 2}^5$

$= - \left[ {\left( {\frac{4}{2} - 4} \right) - \left( {\frac{{25}}{2} - 10} \right)} \right] $ $+ \left[ {\left( {\frac{{25}}{2} + 10} \right) - \left( {\frac{4}{2} - 4} \right)} \right]$

$= - \left( { - 2 - \frac{5}{2}} \right) + \left( {\frac{{45}}{2} + 2} \right)$

$= 2 + \frac{5}{2} + \frac{{45}}{2} + 2$

$ = 4 + 25 = 29$

View full question & answer→

Question 593 Marks

By using the properties of definite integrals, evaluate the integral $\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^5}xdx}}{{{{\sin }^5}x + {{\cos }^5}x}}} $

Answer

Let $I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^5}xdx}}{{{{\sin }^2}x + {{\cos }^5}x}}} dx$…(i)
$ \Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^5}\left( {\frac{\pi }{2} - x} \right)}}{{{{\sin }^5}\left( {\frac{\pi }{2} - x} \right) + {{\cos }^5}\left( {\frac{\pi }{2} - x} \right)}}dx} $
$\left[ {\because \int\limits_0^{{a }{}} {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)} dx } } \right]$
$\Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sin }^5x}}}{{{{\cos }^5}x + {{\sin }^5}x}}dx} $ …(ii)
Adding equations (i) and (ii),
$2I = \int\limits_0^{\frac{\pi }{2}} {\left( {\frac{{{{\cos }^5}x}}{{{{\cos }^5}x + {{\sin }^5}x}} + \frac{{{{\sin }^5}x}}{{{{\cos }^5}x + {{\sin }^5}x}}} \right)dx} $
$= \int\limits_0^{\frac{\pi }{2}} {\left( {\frac{{{{\cos }^5}x + {{\sin }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}} \right)dx} $
$ \Rightarrow 2I = \int\limits_0^{\frac{\pi }{2}} {1dx} $
$\Rightarrow 2I = \frac{\pi }{2}$
$\Rightarrow I = \frac{\pi }{4}$

View full question & answer→

Question 603 Marks

By using the properties of definite integrals, evaluate the integral $\int\limits_0^a {\frac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}} dx$

Answer

Let $I = \int\limits_0^a {\frac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}} dx$…(i)
$\Rightarrow I = \int\limits_0^a {\frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt {a - \left( {a - x} \right)} }}dx} $
$\left[ {\because \int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)dx = } } } \right]$
$= \int\limits_0^a {\frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt x }}} dx$ …(ii)
Adding eq. (i) and (ii),

$2I = \int\limits_0^a {\left( {\frac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }} + \frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt x }}} \right)dx} $
$= \int\limits_0^a {\left( {\frac{{\sqrt x + \sqrt {a - x} }}{{\sqrt x + \sqrt {a - x} }}} \right)dx}$
$= \int\limits_0^a {1dx} = \left( x \right)_0^a = a$
$ \Rightarrow I = \frac{a}{2}$

View full question & answer→

Question 613 Marks

By using the properties of definite integrals, evaluate the integral $\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}} dx$

Answer

Let $I = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}dx} $ …(i)
$ \Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( {\frac{\pi }{2} - x} \right) - \cos \left( {\frac{\pi }{2} - x} \right)}}{{1 + \sin \left( {\frac{\pi }{2} - x} \right)\cos \left( {\frac{\pi }{2} - x} \right)}}} dx$
$= \int\limits_0^{\frac{\pi }{2}} {\frac{{\cos x - \sin x}}{{1 + \cos x\sin x}}} dx$
$= - \int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \cos x\sin x}}} dx$…(ii)
Adding eq. (i) and (ii), we have $2I = 0 \Rightarrow I = 0$

View full question & answer→

Question 623 Marks

By using the properties of definite integral, evaluate the integral $\int_{0}^{\pi} \frac{x d x}{1+\sin x}$

Answer

Given, $\int_{0}^{\pi} \frac{\mathrm{x}}{1+\sin \mathrm{x}} \mathrm{dx}$
Let, $I=\int_{0}^{\pi} \frac{x}{1+\sin x} d x$ .....(i)
as, $\left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\}$
$\Rightarrow \mathrm{I}=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})}{1+\sin (\pi-\mathrm{x})} \mathrm{dx}$
$\Rightarrow I=\int_{0}^{\pi} \frac{(\pi-x)}{1+\sin x} d x$ .....(ii)
Adding (i) and (ii), we get
$2 \mathrm{I}=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})+\mathrm{x}}{1+\sin \mathrm{x}} \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=\int_{0}^{\pi} \frac{\pi}{1+\sin \mathrm{x}} \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=\pi \int_{0}^{\pi} \frac{(1-\sin \mathrm{x})}{(1+\sin \mathrm{x})(1-\sin \mathrm{x})} \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=\pi \int_{0}^{\pi} \frac{(1-\sin \mathrm{x})}{\cos ^{2} \mathrm{x}} \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=\pi \int_{0}^{\pi}\left\{\frac{1}{\cos ^{2} \mathrm{x}}-\frac{\sin \mathrm{x}}{\cos ^{2} \mathrm{x}}\right\} \mathrm{d} \mathrm{x}$
$2 \mathrm{I}=\pi \int_{0}^{\pi}\left\{\sec ^{2} x-\tan x \sec x\right\} d x$
$\Rightarrow 2 \mathrm{I}=\pi[\tan \mathrm{x}-\sec \mathrm{x}]_{0}^{\pi}$
$\Rightarrow 2 \mathrm{I}=\pi[2]$
$\Rightarrow I=\pi$

View full question & answer→

Question 633 Marks

By using the properties of definite integrals, evaluate the integral $\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}xdx} $

Answer

Let $I = \int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}xdx} $…(i)
$= \int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}\left( {\frac{\pi }{2} - x} \right)dx} $
$\left[ {\because \int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)dx = } } } \right]$
$ \Rightarrow I= \int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}xdx} $ …(ii)
Adding equation (i) and (ii),
$2I = \int\limits_0^{\frac{\pi }{2}} {\left( {{{\cos }^2}x + {{\sin }^2}x} \right)dx} $
$= \int\limits_0^{\frac{\pi }{2}} {1dx} $
$\Rightarrow 2I = \left( x \right)_0^{\frac{\pi }{2}}$
$ \Rightarrow 2I = \frac{\pi }{2}$
$ \Rightarrow I = \frac{\pi }{4}$

View full question & answer→

MATHS 3 Marks Question