-0-2cos x-sin x ex d x-A- 0B- 1C- -1D- 2

The correct option is C 1∫_0^π/2[cosx+( sin x)]e^x dx Using the property ∫ e^x(f(x)+f'(x))dx=e^xf(x)+c ∫_0^π/2[cosx+( sin x)]e^x dx =(e^x cos x)_0^π/2= 1 ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

∫0π/2cos x-si...

Question

$\int_{0}^{\frac{π}{2}} (c o s x - s i n x) e^{x} d x =$

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-1

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Solution

The correct option is B -1
$\int_{0}^{\frac{π}{2}} [c o s x + (- s i n x)] e^{x} d x$

Using the property $\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) + c$

$\int_{0}^{\frac{π}{2}} [c o s x + (- s i n x)] e^{x} d x$

$= (e^{x} c o s x)_{0}^{\frac{π}{2}} = - 1$

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∫π20(cos x−sin x)ex dx=

The correct option is B -1∫π20[cosx+(−sin x)]ex dx Using the property ∫ex(f(x)+f′(x))dx=exf(x)+c ∫π20[cosx+(−sin x)]ex dx =(ex cos x)π20=−1

$\int_{0}^{\frac{π}{2}} (c o s x - s i n x) e^{x} d x =$

The correct option is B -1
$\int_{0}^{\frac{π}{2}} [c o s x + (- s i n x)] e^{x} d x$

Using the property $\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) + c$

$\int_{0}^{\frac{π}{2}} [c o s x + (- s i n x)] e^{x} d x$

$= (e^{x} c o s x)_{0}^{\frac{π}{2}} = - 1$