Tag: fundamental theorem of calculus

Questions Related to fundamental theorem of calculus

The value of the integral $\displaystyle\int{\sin{x}{\cos}^{4}{x}dx}$ where $x\in\left[-1,\,1\right]$ is 

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. 1

  2. 1\2

  3. 0

  4. 4

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Correct Option: C
Explanation:
$f\left(x\right)=\sin{x}{\cos}^{4}{x},$

$f\left(-x\right)=\sin{\left(-x\right)}{\cos}^{4}{\left(-x\right)}=-f\left(x\right)$

Since $f\left(x\right)$ is an odd function, $\displaystyle\int _{-1}^{1}\sin{x}{\cos}^{4}{x}dx=0$

If $\Delta (x)=\left| \begin{matrix} 1+x+2{ x }^{ 2 } & x+3 & 1 \ x+2{ x }^{ 2 } & x & 3 \ 3x+6{ x }^{ 2 } & 3x+11 & 9 \end{matrix} \right| $ then $\displaystyle \int^{1} _{0}\Delta (x)dx$ is

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $\dfrac {176}{5}$

  2. $-\dfrac {176}{3}$

  3. $\dfrac {186}{3}$

  4. $-\dfrac {192}{3}$

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Correct Option: A
If $\phi{\left(x\right)}={\phi}^{\prime}{\left(x\right)}$ and $\phi{\left(1\right)}=2$ then $\phi{\left(3\right)}$  is equal to
business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. ${ \phi  }^{ 2 }$

  2. $2{ \phi  }^{ 2 }$

  3. $3{ \phi  }^{ 2 }$

  4. $2{ \phi  }^{ 3 }$

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Correct Option: A
Explanation:
$ \phi(x) = \phi '(x) \Rightarrow  \phi(x) = \frac{2\phi(x)}{dx} $

$ \Rightarrow  $  $\int  dx = \int \frac{d(\phi (x))}{\phi(x)}$

$ \Rightarrow x+c = ln \phi (x) \Rightarrow \phi (x) = k.e^{x}$

$ \phi(1) = 2 \Rightarrow  2 = k.e^{1}  $ $ \Rightarrow k=2e^{-1}$

$  \therefore \phi (x) = 2.e^{x-1}$

$ \phi(3) = 2.e^{2} = \phi^{2}$

$ \phi(3) = \phi^{2}$


$\displaystyle \int _{1}^{4}\frac{\mathrm{x}\mathrm{d}\mathrm{x}}{\sqrt{2+4\mathrm{x}}}=$

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $\displaystyle \frac{1}{2}$

  2. $\displaystyle \frac{1}{\sqrt{2}}$

  3. $\displaystyle \frac{3}{2}$

  4. $\displaystyle \frac{3}{\sqrt{2}}$

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Correct Option: D
Explanation:

$\int _{1}^{4}\dfrac{x   dx}{\sqrt{2 + 4x}}=\dfrac{1}{2}\int _{1}^{4}\dfrac{x   dx}{\sqrt{x+\dfrac{1}{2}}}$
$=\dfrac{1}{2}\left [ \int _{1}^{4}\dfrac{(x+\dfrac{1}{2})dx}{\sqrt{x+\dfrac{1}{2}}}-\int _{1}^{4}\dfrac{\dfrac{1}{2}dx}{\sqrt{x+\dfrac{1}{2}}} \right ]$
$=\dfrac{1}{2}\left [ \int _{1}^{4} \sqrt{x+\dfrac{1}{2}} dx-\int _{1}^{4}(x+\dfrac{1}{2})^{\dfrac{1}{2}} \int _{1}^{4} \right ]$
$=\dfrac{1}{2} \left [ \dfrac{2}{3} (x+\dfrac{1}{2})^{\dfrac{3}{2}} \int _{1}^{4}-(x+\dfrac{1}{2})^{\dfrac{4}{2}} \int _{1}^{4} \right ]$
$=\dfrac{1}{3} \left [ \left ( \dfrac{9}{2} \right )^{\dfrac{3}{2}}-\left ( \dfrac{3}{2} \right )^{\dfrac{3}{2}} \right ] -\dfrac{1}{2} \left [ \left ( \dfrac{9}{2} \right )^{\dfrac{1}{2}}-\left ( \dfrac{3}{2} \right )^{\dfrac{1}{2}} \right ]$
$=\dfrac{1}{3} \left [ \left ( \dfrac{9}{2} \right )\left ( \dfrac{9}{2} \right )^{\dfrac{1}{2}}-\left ( \dfrac{3}{2} \right )\left ( \dfrac{3}{2} \right )^{\dfrac{1}{2}} \right ] - \dfrac{1}{2} \left [ \left ( \dfrac{3}{\sqrt{2}} \right )-\dfrac{\sqrt{3}}{\sqrt{2}} \right ]$
$=\dfrac{3}{\sqrt{2}}$

The value of $\displaystyle \int _{0}^{2}(x-\log _{2}a)dx=2\log _{2}(\frac{2}{a})$ for which of the following conditions?

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $\mathrm{a}>0$

  2. $\mathrm{a}>2$

  3. $\mathrm{a}=4$

  4. $\mathrm{a}=8$

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Correct Option: A
Explanation:

$\int _{ 0 }^{ 2 }{ (x-\log _{ 2 }a } )dx=2\log _{ 2 }(\cfrac { 2 }{ a } )$

The solution exists only if function is defined.
$ x-\log _{ 2 }a\longrightarrow$ defined
$ x\longrightarrow$ is defined for all values 
But $\log _{ 2 }a\longrightarrow$ defined for all values
But $ \log _{ 2 }a\longrightarrow$ defined for only a>0$
$\therefore \log (0)$ and $\log \text {(negative values)} )\longrightarrow$ not defined
Hence, required condition is $a>0$.

Consider the integral $I=\displaystyle\int^{\pi} _0 ln(\sin x)dx$.What is $\displaystyle\int^{\dfrac{\pi}{2}} _{0}$ ln $(\sin x)dx$ equal to?

physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $4I$

  2. $2I$

  3. $I$

  4. $\dfrac{I}{2}$

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Correct Option: D
Explanation:

$I = \displaystyle \int _{0}^{\pi} {ln(\sin x)dx}$

 using property,
$I = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {(ln(\sin (2\pi -x) +ln(\sin x)) dx}$

we know that $\sin(x) = \sin(2\pi -x)$ 

$I = 2\displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin x)dx}$

$\dfrac{I}{2} = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin x)dx}$

Consider the integral $I=\displaystyle\int^{\pi} _0 ln(\sin x)dx$.What is $\displaystyle\int^{\frac{\pi} {2}} _0 ln(\cos x)dx$ equal to?

physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $\dfrac{I}{2}$

  2. $I$

  3. $2I$

  4. $4I$

Show answer
Correct Option: A
Explanation:

$I = \displaystyle \int _{0}^{\pi} {ln(\sin x)dx}$

 using property,
$I = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {(ln(\sin (2\pi -x) +ln(\sin x)) dx}$

we know that $\sin(x) = \sin(2\pi -x)$ 
$I = 2\displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin x)dx}$

$\dfrac{I}{2} = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin x)dx}$
 by property,

$\dfrac{I}{2} = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin (\dfrac{\pi}{2} - x))dx}$

$\dfrac{I}{2} = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\cos x)dx}$

$ \int _{\sin x}^1 t^2 f(t) dt = 1 - \sin x \forall x \epsilon (0, \pi / 2 ) $ then $ f \left( \dfrac {1}{\sqrt3} \right) $ is :

physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $3$

  2. $\sqrt3$

  3. $1/3$

  4. None of these

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Correct Option: A
Explanation:

$ \int _{\sin x}^1 t^2 f(t) dt = 1 - \sin x \forall x \epsilon (0, \pi/2 ) $
Differentiating both sides we get 
$ \dfrac {d}{dx} (1) [ 1 \cdot f (1)] - \cos x ( \sin^2 x) f ( \sin x) = -\cos x  $
$ \Rightarrow f ( \sin x) = \dfrac {1}{ \sin^2 x} $
$ \therefore f \left( \dfrac {1}{\sqrt3} \right) = f \left( \sin \left( \sin^{-1} \dfrac {1}{\sqrt3} \right) \right) $
$ = \left[ \dfrac {1}{ \sin \left( \sin^{-1} \dfrac {1}{\sqrt3} \right)} \right]^2 = 3 $

Consider the integrals ${I _1} = \int _0^1 {{e^{ - x}}{{\cos }^2}xdx,} {I _2} = \int _0^1 {{e^{ - {x^2}}}{{\cos }^2}xdx,} {I _3} = \int _0^1 {{e^{ - x}}dx} $ and ${I _4} = \int _0^1 {{e^{ - (1/2){x^2}}}} dx$. The greatest of these integrals is

physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $I _1$

  2. $I _2$

  3. $I _3$

  4. $I _4$

Show answer
Correct Option: D
Explanation:

$I _1=\int _{0}^{4}e^{-x} cos^2x dx$
$I _2=\int _{0}^{1}e^{-x^2}cos^2x dx$
Both have $cos^2x$ so value get restricted more in (0, 1)
Now in (0, 1) $e^{-\frac {x^2}{2}}>e^{-x}$
$\therefore \int _{0}^{1}e^{-\frac {x^2}{2}}>\int _{0}^{1}e^{-x}$
$\therefore I _4>I _3>I _1>I _2$

Let $ f\left( a,b \right) =\int _{ a }^{ b }{ \left( { x }^{ 2 }-4x+3 \right) dx,\left( b>a \right)  }$ then

physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $ f\left( a,3 \right)$ is least when $a=1$

  2. $f\left( 4,b \right)$ is an increasing function $ \forall b\ge 4$

  3. $ f\left( 0,b \right)$ is least for $b=2$

  4. $ \min { \left{ f\left( a,b \right) \right} =-\dfrac { 4 }{ 3 } } \forall a,b\in R$

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Correct Option: A
Explanation:

First of all integrate the function.

$\displaystyle \int _{  }^{  }{ \left( { x }^{ 2 }-4x+3 \right) dx= } \left[ \frac { { x }^{ 3 } }{ 3 } -4\frac { { x }^{ 2 } }{ 2 } +3x \right] $

Apply the limits,
$\displaystyle f(a,b)={ \left[ \frac { { x }^{ 3 } }{ 3 } -4\frac { { x }^{ 2 } }{ 2 } +3x \right]  } _{ a }^{ b }$

Here $b=3$

Therefore,
$\displaystyle f(a,3)={ \left[ \frac { { x }^{ 3 } }{ 3 } -4\frac { { x }^{ 2 } }{ 2 } +3x \right]  } _{ a }^{ b }=\left[ \frac { 27 }{ 3 } -2\times 9+9 \right] -\left[ \frac { { a }^{ 3 } }{ 3 } -2{ a }^{ 2 }+3a \right] \ \displaystyle =2{ a }^{ 2 }-\frac { { a }^{ 3 } }{ 3 } -3a$

Hence,
Differentiate the above function to find the maximum or minimum.
$\displaystyle { f }^{ \prime  }\left( a,3 \right) =4a-{ a }^{ 2 }-3=0$

therefore $\displaystyle a=1,3$

Check whether the function is minimum or maximum.
$\displaystyle { f }^{ \prime \prime  }\left( a,3 \right) =4-{ 2a }=2$ which  is greater than $0$.
Hence the function $f(a,3)$ is minimum at $a=1$.