-0- sinx dx is equal to -

We know, from second fundamental theorem ∫_a^bf(x)dx=F(b) F(a); where F'(x)=f(x) So,∫_0^πsin x dx = ( cos(π)) ( cos(0)) [Since ∫ sin x dx = cos(x)] = ...

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Standard XI

Mathematics

Second Fundamental Theorem of Calculus

∫0π sinx dx i...

Question

$\int_{0}^{π} s i n x$ dx is equal to -

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Solution

The correct option is C 2
We know, from second fundamental theorem -
$\int_{a}^{b} f (x) d x = F (b) - F (a); w h e r e F^{'} (x) = f (x)$
$S o, \int_{0}^{π} s i n x d x = (- c o s (π)) - (- c o s (0))$ [Since $\int s i n x$ dx = - cos(x)]
= - (-1) - (-1)
= 1 + 1
= 2

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∫π0sinx dx is equal to -

The correct option is C 2We know, from second fundamental theorem - ∫baf(x)dx=F(b)−F(a);whereF′(x)=f(x) So,∫π0sinxdx=(−cos(π))−(−cos(0)) [Since ∫sinx dx = - cos(x)] = - (-1) - (-1) = 1 + 1 = 2

$\int_{0}^{π} s i n x$ dx is equal to -

The correct option is C 2
We know, from second fundamental theorem -
$\int_{a}^{b} f (x) d x = F (b) - F (a); w h e r e F^{'} (x) = f (x)$
$S o, \int_{0}^{π} s i n x d x = (- c o s (π)) - (- c o s (0))$ [Since $\int s i n x$ dx = - cos(x)]
= - (-1) - (-1)
= 1 + 1
= 2