VITEEE 2014 :: Mathematics :: General questions

• Mathematics - General questions

The area bounded by $y=xe^{|x|}$ and lines |x|=1 , y=0 is 

Answer:

Explanation:

|x|=1

$\therefore$   $x\pm 1$

$\therefore$   $y=xe^{|x|}$=$\begin{cases}x.e^{-x},& -1<x<0\\x.e^{x}, & 0<x<1\end{cases}$

$\therefore$  Required area = $|\int_{-1}^{0} x e^{-x}dx+\int_{0}^{1} xe^{x}dx|$

 = $|\left[ -x. e^{-x}-e^{-x}\right]_{-1}^{0}+\left\{ x.e^{x}-e^{x}\right\}^{1}_{0}|$

$=|\left\{(0-1)-(1.e-e)\right\}|+|\left\{(e-e)-(0-1)\right\}|$

 =1+1=2 sq units 

The area in the first quadrant between $x^{2}+y^{2}=\pi^{2}$  and $y= \sin x$ is 

Answer:

Explanation:

 $x^{2}+y^{2}=\pi^{2}$ is a circle of radius $\pi$  and centre at the origin

Required of area=  Area  of circle (1 st quadrant) - $\int_{0}^{\pi} \sin x dx$

 = $\frac{\pi \pi^{2}}{4}-[-\cos x]_{0}^{\pi}=\frac{\pi^{3}}{4}+(\cos \pi-\cos 0)$

 =$ \frac{\pi^{3}}{4}+(-1-1)=\frac{\pi^{3}}{4}-2=\frac{\pi^{3}-8}{4}$

If l(m,n) = $\int_{0}^{1} t^{m}(1+t)^{n} dt, $ then the expression for l(m,n) in terms of l(m+1,n+1) is 

Answer:

Explanation:

$l(m,n)=l=\int_{0}^{1} t^{m}(1+t)^{n} dt$

$\Rightarrow l(m,n)=\left[(1+t)^{n}\frac{t^{m+1}}{m+1}\right]_{0}^{1}$

$-\frac{n}{m+1}\int_{0}^{1} (1+t)^{n-1}.t^{m+1}.dt$

$=\frac{2^{n}}{m+1}-\frac{n}{m+1}l(m+1,n-1)$

The value of $  \int_{0}^{\sqrt{2}} [x^{2}]dx$ , where [.] is the greatest integer function is 

Answer:

Explanation:

$\int_{0}^{\sqrt{2}} \left[x^{2}\right]dx=\int_{0}^{1} \left[x^{2}\right]dx+\int_{1}^{\sqrt{2}}\left[x^{2}\right]dx $

$=\int_{0}^{1} 0dx+\int_{1}^{\sqrt{2}}1dx $

  $= \left[x\right]_{1}^{\sqrt{2}}=\sqrt{2}-1$

$  \int_{0}^{2x} (\sin x+|\sin x|)dx$ is equal to 

Answer:

Explanation:

$  \int_{0}^{2x} (\sin x+|\sin x|)dx$

= $  \int_{0}^{\pi} (\sin x+\sin x)dx$+$  \int_{\pi}^{2 \pi} (\sin x-\sin x)dx$

  =$ 2\int_{0}^{\pi} \sin x dx$+0=$ 2 \int_{0}^{\pi} \cos x dx$

 =    $-2(\cos \pi-\cos 0)=-2(-1-1)=4$

• Mathematics - General questions