Tag: interpretation of define integral as an area

Quizzes (0) Questions (26)

Questions Related to interpretation of define integral as an area

Multiple choice 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

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

  1. 1

  2. 1\2

  3. 0

  4. 4
Reveal answer Fill a bubble to check yourself
C Correct answer
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, $\displaystyle\int _{-1}^{1}\sin{x}{\cos}^{4}{x}dx=0$
Multiple choice 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

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

  1. $\dfrac {176}{5}$
  2. $-\dfrac {176}{3}$
  3. $\dfrac {186}{3}$
  4. $-\dfrac {192}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Expanding the determinant Delta(x) and then integrating term by term from 0 to 1 yields 176/5.

Multiple choice 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
If $\phi{\left(x\right)}={\phi}^{\prime}{\left(x\right)}$ and $\phi{\left(1\right)}=2$ then $\phi{\left(3\right)}$  is equal to
  1. ${ \phi }^{ 2 }$
  2. $2{ \phi }^{ 2 }$
  3. $3{ \phi }^{ 2 }$
  4. $2{ \phi }^{ 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
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}$
Multiple choice 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


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

  1. $\displaystyle \frac{1}{2}$
  2. $\displaystyle \frac{1}{\sqrt{2}}$
  3. $\displaystyle \frac{3}{2}$
  4. $\displaystyle \frac{3}{\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice 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

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

  1. $\mathrm{a}>0$
  2. $\mathrm{a}>2$
  3. $\mathrm{a}=4$
  4. $\mathrm{a}=8$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $a>0$.

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

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?

  1. $4I$
  2. $2I$
  3. $I$
  4. $\dfrac{I}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\sin(x) = \sin(2\pi -x)$ 

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

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

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

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?

  1. $\dfrac{I}{2}$
  2. $I$
  3. $2I$
  4. $4I$
Reveal answer Fill a bubble to check yourself
A Correct answer
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,

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

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

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

$ \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 :

  1. $3$
  2. $\sqrt3$
  3. $1/3$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $

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

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

  1. $I _1$
  2. $I _2$
  3. $I _3$
  4. $I _4$
Reveal answer Fill a bubble to check yourself
D Correct answer
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) $e^{-\frac {x^2}{2}}>e^{-x}$
$\therefore \int _{0}^{1}e^{-\frac {x^2}{2}}>\int _{0}^{1}e^{-x}$
$\therefore I _4>I _3>I _1>I _2$

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

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

  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$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $a=1$.