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Let f(x) = $m i n {[x - 1], [x - 2], . . . ., [x - 10]}$ where [t] denotes the greatest integer $\leq$ t. Then $\int_{0}^{10} f (x) d x + \int_{0}^{10} {(f (x))}^{2} d x + \int_{0}^{10} | f (x) | d x$ is equal to………..

2 Views | Posted 2 months ago

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Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Use [x + n] = n + [x], where $n \in z$ f (x) = [x] – 10 will be minimum for $x \in [0, 10]$ break the limits as G.I.F. is discontinuous at integral points.

$\int_{0}^{10} f (x) d x = - 10 - 9 - . . . . - 1$

$= \frac{- 10.11}{2}$

$\int_{0}^{10} | f (x) | d x = 10 + 9 + . . . . + 1$

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Let f(x) = $m i n {[x - 1], [x - 2], . . . ., [x - 10]}$ where [t] denotes the greatest integer $\leq$ t. Then $\int_{0}^{10} f (x) d x + \int_{0}^{10} {(f (x))}^{2} d x + \int_{0}^{10} | f (x) | d x$ is equal to………..

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Let f(x) = m i n { [ x − 1 ] , [ x − 2 ] , . . . . , [ x − 1 0 ] } where [t] denotes the greatest integer ≤ t. Then ∫ 0 1 0 f ( x ) d x + ∫ 0 1 0 ( f ( x ) ) 2 d x + ∫ 0 1 0 | f ( x ) | d x is equal to………..

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Let f(x) = $m i n {[x - 1], [x - 2], . . . ., [x - 10]}$ where [t] denotes the greatest integer $\leq$ t. Then $\int_{0}^{10} f (x) d x + \int_{0}^{10} {(f (x))}^{2} d x + \int_{0}^{10} | f (x) | d x$ is equal to………..