Mathematics

# A curve is represented parametrically by the equation $x=e^t \cos t$ and $y=e^t \sin t$, where $t$ is a parameter. Then,If $F(t)=\int (x+y)dt$, then the value of $F\left (\displaystyle \frac {\pi}{2}\right )-F(0)$ is

##### ANSWER

$e^{\tfrac{\pi}{2}}$

##### SOLUTION
Given,

$F(t)=\int (x+y)dt$

$=\int (e^t \cos t+e^t \sin t)dt$

$=\dfrac{e^t\sin \left(t\right)}{2}+\dfrac{e^t\cos \left(t\right)}{2}-\dfrac{e^t\cos \left(t\right)}{2}+\dfrac{e^t\sin \left(t\right)}{2}$

$=e^t\sin t+c$

Now,

$F\left ( \dfrac{\pi}{2} \right )-F(0)=e^{\frac{\pi}{2}}\sin \dfrac{\pi}{2}-(e^0\sin 0)$

$=e^{\frac{\pi}{2}}(1)-1 (0)$

$=e^{\frac{\pi}{2}}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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