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Second Fundamental Theorem of Calculus

Let fx = 1/2 ...

Question

Let $f (x) = \frac{1}{2} a_{0} + \sum_{i - 1}^{n} a_{i} c o s (i x) + \sum_{j - 1}^{n} b_{j} s i n (j x), t h e n \int_{- π}^{π} f (x) c o s k x d x$ then is equal to

A

a_{k}

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B

b_{k}

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C

π a_{k}

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D

π b_{k}

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Solution

The correct option is C $π a_{k}$
Conceptual

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Similar questions

\sum_{j = 1}^{n} \sum_{i = 1}^{n} i

f (x) = {sin}^{- 1} x + {cos}^{- 1} x,

\frac{π}{2}

f (x) = \frac{1}{2} a_{0} + \sum_{i - 1}^{n} a_{i} c o s (i x) + \sum_{j - 1}^{n} b_{j} s i n (j x), t h e n \int_{- π}^{π} f (x) c o s k x d x

then is equal to

f (x) = sin \sqrt{p} x,

p = [a] =

the greatest integer less than or equal to

a

. If the period of

f (x)

π

f (x) = {(sin ({tan}^{- 1} x) + sin ({cot}^{- 1} x))}^{2} - 1

| x | > 1

\frac{d y}{d x} = \frac{1}{2} \frac{d}{d x} ({sin}^{- 1} (f (x)))

y (\sqrt{3}) = \frac{π}{6}

y (- \sqrt{3})

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Second Fundamental Theorem of Calculus

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