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Integrals Test - 55

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Integrals Test - 55
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  • Question 1
    1 / -0
     If $$ b > a$$ and $$ \displaystyle I = \int_{a}^{b}\frac{dx}{\sqrt{ (x-a)(b-x)}}$$ then $$I$$ equals
    Solution
    Substitute $$\displaystyle t=\frac{1}{2} ( x- a+x-b)=x- \frac{1}{2}(a + b),$$ so that
    $$(x-a)(b-x)=(t+c)(c-t)=c^{2}-t^{2}$$
    where $$\displaystyle  c = \frac{ 1}{2} (b -a).$$
    Thus,
    $$\displaystyle I=\int_{-c}^{c} \frac{dx}{\sqrt{(c^{2}-t^{2})}}$$
    $$\displaystyle =2 \int_{0}^{c} \frac{dx}{\sqrt{ (c^{2}-t^{2})}}=2 \sin^{-1} \left.\left( \frac{t}{c}\right) \right ]_{0}^{c}$$
    $$ =2 [\sin^{-1}(1)-0]=\pi$$
  • Question 2
    1 / -0
    A function $$f $$ is defined by $$\displaystyle f(x)=\frac{1}{2^{r-1}},\frac{1}{2r}<x\leq \frac{1}{2^{r-1}},r=1,2,3,.....$$ then the value of $$ \displaystyle \int _{0}^{1}f(x)dx $$ is equal
    Solution
    $$\displaystyle \int_{0}^{1}f(x) dx =\sum_{r=1}^{\infty} \int_{2^{-r}}^{2^{-(r-1)}} \frac{1}{2^{r-1}}dx $$
    $$\displaystyle =\sum_{r=1}^{\infty}\frac{1}{2^{r-1}}[2^{-(r-1)}-2^{-r}]$$
    $$\displaystyle =\sum_{1}^{\infty}2^{-2(r-1)}- \sum _{1}^{\infty} 2^{-2r+1}$$
    $$\displaystyle =(2^{2}-2)\sum_{1}^{\infty}2^{-2r}=2\cdot \frac{1}{4}\cdot \frac{1}{1-1/4}=\frac{2}{3}\cdot $$
  • Question 3
    1 / -0
    Evaluate $$\displaystyle \int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^{2}+x^{3}}}$$
    Solution
    $$\displaystyle l=\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{dx}{\sqrt{x+x^{2}+x^{3}}}$$
    $$\displaystyle =\int_{0}^{1}\frac{\left ( 1-x^{2} \right )}{\left ( x^{2}+2x+1 \right )\sqrt{x+x^{2}+x^{3}}}$$
    $$\displaystyle =\int_{0}^{1}\frac{\frac{1}{x^{2}}-1dx}{\left ( x+\frac{1}{x}+2 \right )\sqrt{x+\frac{1}{x}+1}}$$
    Put $$\displaystyle x+\frac{1}{x}+1=t^{2}\Rightarrow \left ( 1-\frac{1}{x^{2}}dx \right )=2tdt$$
    $$\displaystyle l=\int_{\infty }^{\sqrt{3}}\frac{-2tdt}{\left ( t^{2}+1 \right )t}=2\int_{\sqrt{3}}^{\infty }\frac{dt}{t^{2}+1}=2\left [ \tan^{-1} \right ]^{\infty }_{\sqrt{3}}=\frac{\pi }{3}$$
  • Question 4
    1 / -0
    $$\displaystyle \int_{e^{e^{e}}}^{e^{e^{e^{e}}}}\frac{dx}{xlnx\cdot ln\left ( lnx \right )\cdot ln\left ( ln\left ( lnx \right ) \right )}$$ equals
    Solution
    Let $$\displaystyle I=\int _{ { e }_{ { e }_{ e } } }^{ { e }^{ { e }^{ { e }^{ e } } } }{ \frac { 1 }{ x\log { x } .\log { \left( \log { x }  \right) . } \log { \left( \log { \left( \log { x }  \right)  }  \right)  }  }  } dx$$

    Substitute $$\displaystyle \log { \left( \log { x }  \right)  } =t\Rightarrow \frac { 1 }{ x\log { x }  } dx=dt$$

    $$\displaystyle I=\int _{ e }^{ { { e }^{ e } } }{ \frac { 1 }{ t\log { t }  } dt } =\left[ \log { \log { t }  }  \right] _{ e }^{ { e }^{ e } }=\left[ 1-0 \right] =1$$
  • Question 5
    1 / -0
    The value of integral $$\displaystyle \int _{ \frac { \pi  }{ 6 }  }^{ \frac { \pi  }{ 3 }  }{ \cfrac { \sin { x } -x\cos { x }  }{ x(x+\sin { x } ) }  } dx$$ is
    Solution
    Let $$I=\displaystyle \int _{ \frac { \pi  }{ 6 }  }^{ \frac { \pi  }{ 3 }  }{ \cfrac { \sin { x } -x\cos { x }  }{ x(x+\sin { x } ) }  } dx$$

    $$\Rightarrow I=\displaystyle \int _{ \frac { \pi  }{ 6 }  }^{ \frac { \pi  }{ 3 }  }{ \cfrac { (x+\sin { x) } -x(1+\cos { x) }  }{ x(x+\sin { x } ) }  } dx$$

    $$\Rightarrow I=\displaystyle \int _{ \frac { \pi  }{ 6 }  }^{ \frac { \pi  }{ 3 }  }{ \left( \cfrac { 1 }{ x } -\cfrac { 1+\cos { x }  }{ x+\sin { x }  }  \right)  } dx$$

    $$\Rightarrow I={ \left[ \log { x }  \right]  }_{ \pi /6 }^{ \pi /3 }-\displaystyle \int _{ \frac { \pi  }{ 6 }  }^{ \frac { \pi  }{ 3 }  }{ \cfrac { 1+\cos { x }  }{ x+\sin { x }  }  } dx$$

    Put $$t=x+\sin { x } $$; $$dt=(1+\cos { x } )dx$$ in $$2^{nd}$$ term

    $$\Rightarrow I=\left( \log { \dfrac { \pi  }{ 3 }  } -\log { \dfrac { \pi  }{ 6 }  }  \right) -\displaystyle \int _{ \left( \frac { \pi  }{ 6 } +\frac { 1 }{ 2 }  \right)  }^{ \left( \frac { \pi  }{ 3 } +\frac { \sqrt { 3 }  }{ 2 }  \right)  }{ \cfrac { dt }{ t }  } $$

    $$\Rightarrow I=\log { 2 } -{ \left[ \log { t }  \right]  }_{ \left( \frac { \pi  }{ 6 } +\frac { 1 }{ 2 }  \right)  }^{ \left( \frac { \pi  }{ 3 } +\frac { \sqrt { 3 }  }{ 2 }  \right)  }$$

    $$\Rightarrow I=\log { 2 } -\left[ \log { \left( \cfrac { \pi  }{ 3 } +\cfrac { \sqrt { 3 }  }{ 2 }  \right)  } -\log { \left( \cfrac { \pi  }{ 6 } +\cfrac { 1 }{ 2 }  \right)  }  \right] $$

    $$\Rightarrow I=\log { 2 } -\log { \left( \cfrac { 2\pi +3\sqrt { 3 }  }{ \pi +3 }  \right)  } \left( \because \log { m } -\log { n } =\log { \cfrac { m }{ n }  }  \right) $$

    $$I=\log { \left( \cfrac { 2(\pi +3) }{ 2\pi +3\sqrt { 3 }  }  \right)  } =\log { \left( \cfrac { 2\pi +6 }{ 2\pi +3\sqrt { 3 }  }  \right)  } $$
  • Question 6
    1 / -0
    Evaluate $$\displaystyle \int_{0}^{1}\left ( tx+1-x \right )^{n}dx,$$ where n is a positive integer and t is a parameter independent of x. Hence $$\displaystyle \int_{0}^{1}x^{k}\left ( 1-x \right )^{n-k}dx=\frac{P}{\left [ ^{n}C_{k}\left ( n+1 \right ) \right ]}for\:k=0,1,......n$$, then $$P=$$
    Solution
    Let $$\displaystyle I=\int _{ 0 }^{ 1 }{ { \left( tx+1-x \right)  }^{ n } } dx=\int _{ 0 }^{ 1 }{ { \left\{ \left( t-1 \right) x+1 \right\}  }^{ n } } dx=\left[ \frac { \left( t-1 \right) { x+1 }^{ n+1 } }{ \left( n+1 \right) \left( t-1 \right)  }  \right] _{ 0 }^{ 1 }$$
    $$\displaystyle =\frac { 1 }{ n+1 } \left( \frac { { t }^{ n+1 }-1 }{ t-1 }  \right) =\frac { 1 }{ n+1 } \left( 1+t+{ t }^{ 2 }+...{ t }^{ n } \right) $$   ...(1)
    Again $$\displaystyle\int _{ 0 }^{ 1 }{ { \left( tx+1-x \right)  }^{ n } } dx=\int _{ 0 }^{ 1 }{ { \left[ \left( 1-x \right) +tx \right]  }^{ n } } dx$$
    $$\displaystyle=\int _{ 0 }^{ 1 }{ \left[ _{  }^{ n }{ { C }_{ 0 } }{ \left( 1-x \right)  }^{ n }+^{ n }{ { C }_{ 1 } }{ \left( 1-x \right)  }^{ n-1 }\left( tx \right) +^{ n }{ { C }_{ 2 } }{ \left( 1-x \right)  }^{ n-2 }{ \left( tx \right)  }^{ 2 }+...+^{ n }{ { C }_{ n } }{ \left( tx \right)  }^{ n } \right]  } dx$$
    $$\displaystyle=\int _{ 0 }^{ 1 }{ \sum _{ r=1 }^{ n }{ ^{ n }{ { C }_{ r } } }  } { \left( 1-x \right)  }^{ n-r }{ \left( tx \right)  }^{ r }dx=\sum _{ r=1 }^{ n }{ ^{ n }{ { C }_{ r } } } \left[ \int _{ 0 }^{ 1 }{ { \left( 1-x \right)  }^{ n-r }{ x }^{ r }dx }  \right] { t }^{ r }$$   ...(2)
    From (1) and (2)
    $$\displaystyle \sum _{ r=1 }^{ n }{ ^{ n }{ { C }_{ r } } } \left[ \int _{ 0 }^{ 1 }{ { \left( 1-x \right)  }^{ n-r }.{ x }^{ r }dx }  \right] { t }^{ r }=\frac { 1 }{ n+1 } \left( 1+t+...{ t }^{ n } \right) $$
    On equating coefficient of $${ t }^{ k }$$ on both sides , we get
    $$\displaystyle ^{ n }{ { C }_{ k } }\left[ \int _{ 0 }^{ 1 }{ { \left( 1-x \right)  }^{ n-r },{ x }^{ r } } dx \right] \Rightarrow \int _{ 0 }^{ 1 }{ { \left( 1-x \right)  }^{ n-k } } { x }^{ h }dx=\frac { 1 }{ { \left( n+1 \right)  }^{ n }{ { C }_{ k } } } $$
  • Question 7
    1 / -0
    Evaluate: $$\displaystyle \int_{0}^{1}\dfrac{1-x}{1+x}\cdot \dfrac{dx}{\sqrt{x+x^{2}+x^{3}}}$$
    Solution
    Consider, $$\displaystyle I=\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{dx}{\sqrt{x+x^{2}+x^{3}}}$$

    $$\displaystyle I=\int_{0}^{1}\frac{\left ( 1-x^{2} \right )}{\left ( x^{2}+2x+1 \right )\sqrt{x+x^{2}+x^{3}}}$$

    $$\displaystyle I=\int_{0}^{1}\frac{\frac{1}{x^{2}}-1dx}{\left ( x+\frac{1}{x}+2 \right )\sqrt{x+\frac{1}{x}+1}}$$

    Put $$\displaystyle x+\frac{1}{x}+1=t^{2}\Rightarrow \left ( 1-\frac{1}{x^{2}}dx \right )=2tdt$$

    So, $$x\rightarrow0 \ \Rightarrow t\rightarrow\infty$$ and, $$x\rightarrow 1 \ \Rightarrow t\rightarrow\sqrt3$$

    $$\displaystyle \Rightarrow I=\int_{\infty }^{\sqrt{3}}\frac{-2tdt}{\left ( t^{2}+1 \right )t}=2\int_{\sqrt{3}}^{\infty }\frac{dt}{t^{2}+1}=2\left [ \tan^{-1} \right ]^{\infty }_{\sqrt{3}}=\frac{\pi }{3}$$
  • Question 8
    1 / -0
    Evaluate $$\displaystyle \int_{0}^{\pi /4}\frac{\cos x-\sin x}{10+\sin 2x}dx$$
    Solution
    Let $$\displaystyle I=\int _{ 0 }^{ \frac { \pi  }{ 4 }  }{ \frac { \cos { x } -\sin { x }  }{ 10+\sin { 2x }  }  } dx$$

    Put $$\\ \cos { x } +\sin { x } =t\Rightarrow \left( -\sin { x } +\cos { x }  \right) dx=dt$$
    $$\displaystyle I=\int _{ 1 }^{ \sqrt { 2 }  }{ \frac { 1 }{ 10+\left( { t }^{ 2 }-1 \right)  } dt } =\int _{ 1 }^{ \sqrt { 2 }  }{ \frac { 1 }{ { t }^{ 2 }+9 }  } dt$$

    $$\displaystyle =\frac { 1 }{ 3 } \left[ \tan ^{ -1 }{ \frac { t }{ 3 }  }  \right] _{ 1 }^{ \sqrt { 2 }  }=\frac { 1 }{ 3 } \left( \tan ^{ -1 }{ \frac { \sqrt { 2 }  }{ 3 } -\tan ^{ -1 }{ \frac { 1 }{ 3 }  }  }  \right) $$

  • Question 9
    1 / -0
    The value of $$\displaystyle \int_{3}^{4}\sqrt {(4 - x)(x - 3)}dx$$ is
    Solution
    Let 
    $$\displaystyle I = \int_{3}^{4} \sqrt {(4 - x)(x - 3)}\ dx$$

    $$\displaystyle = \int_{3}^{4} \sqrt {-x^{2} + 7x - 12}\ dx$$

    $$\displaystyle = \int_{3}^{4} \sqrt {-\left (x^{2} - 7x + \dfrac {49}{4} - \dfrac {49}{4}\right ) - 12}\ dx$$

    $$\displaystyle = \int_{3}^{4} \sqrt {- \left (x - \dfrac {7}{2}\right )^{2} + \dfrac {49}{4} - 12}\ dx$$

    $$\displaystyle = \int_{3}^{4} \sqrt {\dfrac {1}{4} - \left (x - \dfrac {7}{2}\right )^{2}}\ dx$$

    Let 
    $$t = x - \dfrac {7}{2}\rightarrow dt = dx$$

    $$\therefore$$ Upper limit $$= \dfrac {1}{2}$$ and lower limit $$= -\dfrac {1}{2}$$
    $$\displaystyle = \int_{-1/2}^{1/2} \sqrt {\left (\dfrac {1}{2}\right )^{2} - t^{2}}dt$$

    $$\displaystyle = 2\int_{0}^{1/2} \sqrt {\left (\dfrac {1}{2}\right )^{2} - t^{2}}dx$$

    $$\displaystyle = 2\left [\dfrac {t}{2}\sqrt {\dfrac {1}{4} - t^{2}} + \dfrac {1}{8} \sin^{-1} 2t\right ]_{0}^{\frac12}$$

    $$= 2\left [0 + \dfrac {1}{8}\times \dfrac {\pi}{2}\right ]$$

    $$= \dfrac {\pi}{8}$$
  • Question 10
    1 / -0
    The value of the integral $$\displaystyle \int_{\frac {1}{3}}^1\frac {(x-x^3)^{\frac {1}{3}}}{x^4}dx$$
    Solution
    Let $$\displaystyle I=\int_{ {1}/{3}}^1\frac {(x-x^3)^{\frac {1}{3}}}{x^4}dx$$

    Put $$x=\sin\theta $$
    $$\Rightarrow dx=\cos\theta \, d\theta $$

    When $$x=\cfrac {1}{3},\theta =\sin^{-1}\left (\cfrac {1}{3}\right )$$ and when $$x=1, \theta =\cfrac {\pi}{2}$$

    $$\Rightarrow \displaystyle  I=\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\sin\theta-\sin^3\theta)^{\frac {1}{3}}}{\sin^4\theta}cos \theta \,d\theta$$

    $$\displaystyle =\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\sin\theta)^{\frac {1}{3}}(1-\sin^2\theta)^{\frac {1}{3}}}{\sin^4\theta}\cos \theta \, d\theta$$

    $$\displaystyle =\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\sin\theta)^{\frac {1}{3}}(\cos\theta)^{\frac {2}{3}}}{\sin^4\theta}\cos \theta \, d\theta$$

    $$\displaystyle =\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\sin\theta)^{\frac {1}{3}}(\cos\theta)^{\frac {2}{3}}}{\sin^2\theta \sin^2\theta}\cos \theta d\theta$$

    $$\displaystyle =\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\cos\theta)^{\frac {5}{3}}}{(\sin\theta)^{\frac {5}{3}}}cosec^2\theta d\theta$$

    $$\displaystyle=\int_{sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}(\cot\theta)^{\frac {5}{3}}cosec^2\theta d\theta$$

    Put $$\cot \theta=t $$
    $$\Rightarrow -cosec^2\theta \,d\theta=dt$$
    When $$\theta=\sin^{-1}\left (\dfrac {1}{3}\right ),t=2\sqrt 2$$ and when $$\theta=\dfrac {\pi}{2}, t=0$$

    $$\therefore I=-\int_{2\sqrt 2}^0(t)^{\frac {5}{3}}dt$$

    $$\Rightarrow  \displaystyle  I=\int_0^{2\sqrt 2} (t)^{\frac {5}{3}}dt$$

    $$\displaystyle =\left [\frac {3}{8}(t)^{\frac {8}{3}}\right ]_0^{2\sqrt 2}$$

    $$\displaystyle =\frac {3}{8}[(2\sqrt 2)^{\frac {8}{3}}]$$

    $$=\dfrac {3}{8}[(\sqrt 8)^{\frac {8}{3}}]$$

    $$=\cfrac {3}{8}[(8)^{\frac {4}{3}}]$$

    $$=\cfrac {3}{8}[16]$$

    $$=3\times 2$$
    $$=6$$
    Hence, the correct Answer is A.
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