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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 definite integral $\int _{ 0 }^{ \pi /2 }{ \sin { x } \sin { 2x } \sin { 3x } dx } $ is equal to:

  1. $\cfrac{1}{3}$
  2. $-\cfrac{2}{3}$
  3. $-\cfrac{1}{3}$
  4. $\cfrac{1}{6}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\int _0^{\pi/2}\sin x\sin 2x\sin 3x dx$


$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2}2\sin x\sin 2x\sin 3x dx$

$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2}2\sin x\sin 3x\sin 2x dx$

We know that       $2\sin A \sin B=\cos(A-B)-\cos (A+B)$

$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2}(\cos (x-3x)-\cos(x+3x))\sin 2x dx$

$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2}(\cos 2x-\cos 4x)\sin 2x dx$

$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2} \sin 2x \cos 2x dx-\dfrac{1}{2}\int _0^{\pi /2}\sin 2x \cos 4x dx$

$\Rightarrow \dfrac{1}{4}\int _0^{\pi/2} 2\sin 2x \cos 2x dx-\dfrac{1}{4}\int _0^{\pi /2}2\sin 2x \cos 4x dx$

We know that   $2\sin A \cos B=\sin(A+B)+\sin (A-B)$

$\Rightarrow \dfrac{1}{4}\int _0^{\pi/2} (\sin (2x+2x)+\sin (2x-2x))dx-\dfrac{1}{4}\int _0^{\pi /2}(\sin (2x+4x)+\sin (2x-4x)) dx$

$\Rightarrow \dfrac{1}{4}\int _0^{\pi/2} \sin 4xdx-\dfrac{1}{4}\int _0^{\pi /2}(\sin 6x-\sin 2x) dx$

$\Rightarrow \dfrac{1}{4}\int _0^{\pi/2} \sin 4xdx-\dfrac{1}{4}\int _0^{\pi /2}\sin 6x dx+\dfrac{1}{4}\int _{0}^{\pi/4}\sin 2x dx$

$\Rightarrow \dfrac{1}{4}[\dfrac{\sin 4x}{4}] _0^{\pi/4}-\dfrac{1}{4}[\dfrac{-\sin 6x}{6}] _0^{\pi/4}+\dfrac{1}{4}[\dfrac{-\cos 2x}{2}] _0^{\pi/4}$

$\Rightarrow \dfrac{-1}{12}+\dfrac{1}{4}=\dfrac{1}{6}$

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 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

Evaluate $I = \displaystyle \int _{\pi /6}^{\pi /3}\sin x:dx$

  1. $\displaystyle \frac{1-\sqrt{3}}{2}$
  2. $\displaystyle \frac{\sqrt{3}+1}{2}$
  3. $\displaystyle \frac{\sqrt{3}-1}{2\sqrt{3}}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given : $I = \displaystyle \int _{\pi /6}^{\pi /3}\sin x\:dx$

Integeration of $\sin x dx$ is $-cos x dx + c$

$I = -cos x dx$

Substuting the upper and lower limit values we get,

$I = -cos\dfrac{\pi}{3}+cos\dfrac{\pi}{6}$

$I = \dfrac{-\sqrt{3}}{2} + \dfrac{1}{2}$

$I = \dfrac{1-\sqrt{3}}{2}$
Multiple choice business maths commission,brokerage and discount advantages of bill of exchange definition, characteristics and parties of bills of exchange simple transactions related to bills of exchange

A shopkeeper may sell an article for less than the marked price for some reason. The difference between the marked price and the lowered selling price is called the ________.

  1. Commission

  2. Discount

  3. Selling price

  4. Cost
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

discount

Multiple choice business maths commission,brokerage and discount advantages of bill of exchange definition, characteristics and parties of bills of exchange simple transactions related to bills of exchange

Tarun got $30%$ concession on the labelled price of an article and sold it for $Rs. 8750$ with $25%$ profit on the price he bought. What was the labelled price ?

  1. $Rs. 10,000$
  2. $Rs. 12,000$
  3. $Rs. 16,000$

  4. Data inadequate

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let the labelled price be $x$ Rs
$\therefore $ Price at which $=x-\dfrac{30}{10}\times x$
he bought $=\left(\dfrac{7x}{10}\right)$ Rs
Also profit is $25\%$
$\therefore  125\% $ of $\dfrac{7x}{10}=8750$ Rs
$\therefore \dfrac{5}{4}\times \dfrac{7x}{10}=8750$ Rs
$\therefore x=\dfrac{8750\times 4\times 10}{5\times 7}$
$\therefore x=10000$ Rs
$\therefore $ Labelled price is $10000$ Rs