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Let a be a positive real number such that \int_{0}^{a} e^{x-[x]} d x=10 e-9 where [x] is the greatest integer less than or equal to x. Then a is equal to  
Option: 1 \\10-\log _{\mathrm{e}}(1+\mathrm{e})\\
Option: 2 \\10+\log _{\mathrm{e}} 2\\
Option: 3 \\10+\log _{\mathrm{e}} 3\\
Option: 4 \\10+\log _{e}(1+e)

\begin{aligned} & \int_{0}^{a} e^{x-[x]} d x \\ =& \int_{0}^{a} e^{\{x\}} d x . \end{aligned}

Let,  a=[a]+\{a\}

=\int_{0}^{[a]} e^{\{x\}} d x+\int_{[a]}^{[a]+\{a\}} e^{\{x\}} d x .

Now {x} is periodic with period 1 \Rightarrow e^{x} is periodic with period 1.

\begin{aligned} &=[a] \int_{0}^{1} e^{\{ x\}} d x+\int_{0}^{\{a\}} e^{\{x\}} dx\,\, \text{(Properties of Periodic functions)}\\ &\text { In } 0 \leq x<1,\{x\}=x \\ &=[a] \int_{0}^{1} e^{x} d x+\int_{0}^{\{a\}} e^{x} d x \\ &=\left.[a]\left(e^{x}\right)\right|_{0} ^{1}+\left.e^{x}\right|_{0} ^{\{a\}} \end{aligned}

\begin{aligned} &=[a](e-1)+e^{\{a\}}-1 \\ &=[a] e+\left(-[a]+e^{\{a\}}-1\right) \end{aligned}

Given that this equals 10 e-9

\Rightarrow[a] e+\left(e^{\{a\}}-[a]-1\right)=10 e-9

Comparing  \Rightarrow[a]=10 \text { and } e^{\{a\}}-10-1=-9 \Rightarrow\{a\}=\ln 2

\Rightarrow a= [a]+\{a\}=10+\ln (2)

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Kuldeep Maurya

Find the value of \int _{0}^{1}\tan ^{-1}\left ( \frac{1-2x}{1+x-x^{2}} \right )dx

 

 

 
 
 
 
 

\\ \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right)=\tan ^{-1}\left(\frac{x-(1-x)}{1+x(1-x)}\right)\\ \because \tan ^{-1} A-\tan ^{-1} B=\tan ^{-1}\left(\frac{A-B}{1+A B}\right) \\ \tan ^{-1} x-\tan ^{-1}(1-x)= \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right)\\ $Now $ I = \int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x=\int_{0}^{1} \tan ^{-1} x d x-\int_{0}^{1} \tan ^{-1}(1-x) d x \\

\\ \because \int_0^a f(x)dx = \int_0^a f(a-x)dx \\ I=\int_{0}^{1} \tan ^{-1}(1-x) d x-\int_{0}^{1} \tan ^{-1}(x) d x \\ I=-\left(\int_{0}^{1} \tan ^{-1} x d x-\int_{0}^{1} \tan ^{-1}(1-x) d x\right) \\ I=-I \\ 2I=0 \Rightarrow I =0

\\ I=\int_{0}^{1} \tan ^{-1}(1-x) d x-\int_{0}^{1} \tan ^{-1}(x) d x \\ I=-\left(\int_{0}^{1} \tan ^{-1} x d x-\int_{0}^{1} \tan ^{-1}(1-x) d x\right) \\ I=-1 \\ -2I=0

 

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Safeer PP

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Find \int \frac{x+1}{(x+2)(x+3)}dx

 

\\\int \frac{x+1}{(x+2)(x+3)}dx \\\\ =\int \frac{x+2-1}{(x+2)(x+3)})dx \\\\ =\int (\frac{1}{x+3}- \frac{1}{(x+2)(x+3)})dx \\\\ =\int (\frac{1}{x+3}- (\frac{1}{x+2} -\frac{1}{x+3}))dx \\\\ =\int (\frac{2}{x+3}- \frac{1}{x+2} )dx \\\\ = 2\log (x+3) - \log(x+2) + c

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Safeer PP

Evaluate \int \frac{x \sin ^{-1}(x^{2})}{\sqrt{1-x^{4}}}dx

 

 

\\\int \frac{x \sin ^{-1}(x^{2})}{\sqrt{1-x^{4}}}dx \\\\ $ Put $ x^2 = t \\ xdx = \frac{1}{2}dt \\ = \int \frac{1}{2}\frac{ \sin ^{-1}(t)}{\sqrt{1-t^{2}}}dt \\\\ $ Put $ \sin ^{-1}(t) = u \\ \frac{1}{\sqrt{1-t^2}}dt = du \\\\ = \int \frac{1}{2} u du \\\\ = \frac{u^2}{4}+c

\\= \frac{(\sin ^{-1}x^2)^2}{4}+c

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Safeer PP

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Find \int \frac{x+1}{x(1-2x)}dx

 

\int \frac{x+1}{x(1-2x)}dx = \int (\frac{3x}{x(1-2x)} + \frac{1-2x}{x(1-2x)})dx \\\\ = \int \frac{3}{(1-2x)} dx+ \int \frac{1}{x}dx \\\\ = \frac{-3}{2} \log (1-2x) + \log x + c \\\\

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Safeer PP

Evaluate the following integral as the limit of sums \int_{1}^{4}(x^{2}-x)dx.

 

 

\\ \text { Given :} f(x)=\left(x^{2}-x\right) a = 1, b=4, h = \frac{b-a}{n} , n h=3 \\ \int_{1}^{4}\left(x^{2}-x\right) d x=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+f(1+2 h)+\ldots . f(1+\overline{n-1} h)] \\ \\ \int_{1}^{4}\left(x^{2}-x\right) d x = \lim _{h \rightarrow 0} h\left[(1-1)+\left(1+h^{2}+2 h-h-1\right)+\left(1+4 h^{2}+4 h-2 h-1\right)+\ldots+\left(1+(n-1)^{2} h^{2}+2(n-1) h-(n-1) h-1\right)\right] \\ =\lim _{h \rightarrow 0} h\left[h^{2}\left(1^{2}+2^{2}+3^{2}+\ldots \ldots+(n-1)^{2}\right)+h(1+2+3+\ldots+(n-1)]\right] \\ =\lim _{h \rightarrow 0}\left[\frac{n h(n h-h)(2 n h-h)}{6}+\frac{(n h(n h-h))}{2}\right] \\$Put value of nh = 3 $ \\ =9+\frac{9}{2}\\=\frac{27}{2}

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Safeer PP

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Using integration find the area of the region bounded between the two circlesx^{2}+y^{2}=9 and (x-3)^{2}+y^{2}=9

 

 

\\ \text { Point of intersection } \\ x^{2}+y^{2}=9....(i) \\ (x-3)^{2}+y^{2}=9 .....(ii)\\ $Equation (ii) -Equation (i) $ \\(x-3)^{2}-x^{2}=0 \\ (x-3-x)(x-3+x)=0 \\ x = \frac{3}{2}

x - cordinate of A = 3/2 

\\ \text { Required area = Area of OAB + Area of OCB = } 2 \times \text{Area of OAB } \\ = 2 [$ Area between OA and x -axis + Area between AB and x-axis $] \\ = 2 \left[\int_{0}^{\frac{3}{2}} \sqrt{9-(x-3)^{2}} d x+\int_{\frac{3}{2}}^{3} \sqrt{9-x^{2}} d x\right] \\ =4\left[\int_{\frac{3}{2}}^{3} \sqrt{9-x^{2}} d x\right] \\ =4\left[\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right]_{\frac{3}{2}}^{3} \\=4 \left(0+\frac{9\pi}{4} - \frac{9 \sqrt{3}}{8}-\frac{3\pi}{4} \right) \\= \left(6 \pi-\frac{9 \sqrt{3}}{2}\right)

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Safeer PP

Evaluate \int _{1}^{2}\left [ \frac{1}{x}-\frac{1}{2x^{2}} \right ]e^{2x}dx

 

 

\int _{1}^{2}\left [ \frac{1}{x}-\frac{1}{2x^{2}} \right ]e^{2x}dx

\\\text { Put } 2 x=t \\ d x=\frac{1}{2} d t \\\\ I=\int_{1}^{2}\left[\frac{1}{x}-\frac{1}{2 x^{2}}\right] e^{2 x} d x \\\\ =\int_{2}^{4}\left[\frac{1}{t}-\frac{1}{t^{2}}\right] e^{t} d t \\\\ =\left[\frac{1}{t} e^{t}\right]_{2}^{4} \because \int (f(x)+f'(x))e^xdx = e^x +c \\\\ =\frac{e^{4}}{4}-\frac{e^{2}}{2}

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Safeer PP

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Find the value of \int_{0}^{1}x(1-x)^{n}dx

 

\\ \int_{0}^{1} x(1-x)^{n} d x \\\\ =\int_{0}^{1}(1-x)(1-1+x)^{n} d x \because \int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx \\\\ =\int_{0}^{1}\left(x^{n}-x^{n+1}\right) d x \\\\ =\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_{0}^{1} \\\\ =\frac{1}{n+1}-\frac{1}{n+2}\\\\ =\frac{1}{(n+1)(n+2)}

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Safeer PP

Find \int \frac{ x}{x ^{2 }+ 3x + 2} dx.

 

 

 
 
 
 
 

\\ \int \frac{x}{x^{2}+3 x+2} d x \\\\ =\int \frac{x}{(x+1)(x+2)} d x \\\\=\int \frac{2(x+1)-(x+2)}{(x+1)(x+2)} d x \\\\=\int\left(\frac{2}{x+2}-\frac{1}{x+1}\right) d x \\ =2 \log |x+2|-\log |x+1|+C

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Safeer PP

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