Question

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 $F\left(t\right)=\int (x+y)dt$, then the value of $F\left(\frac{\pi }{2}\right)-F\left(0\right)$ is

• A. $1$
• B. $-1$
• C. $e^{\frac{\pi }{2}}$
• D. $0$

Answer 1

The correct option is C $e^{\frac{\pi }{2}}$

$x=e^t\cos t$ and $y=e^t\sin t$
$F\left(t\right)=\int (x+y)dt$

We know that, $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)+c$

In the given integral, $f\left(x\right)=\sin x$
$\therefore F\left(t\right)=\int e^t\left(\cos t+\sin t\right)dt=e^t\sin t+c$
$F\left(\frac{\pi }{2}\right)-F\left(0\right)=\left(e^{\frac{\pi }{2}}\right)\sin \frac{\pi }{2}-e^0\sin 0=e^{\frac{\pi }{2}}-0=e^{\frac{\pi }{2}}$

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