JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Definite Integrals

Let

f (x) = min {[x - 1], [x - 2], \dots, [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 ____.

Answer: 385

Solution

For

x \in [k, k + 1)

where

k = 0, 1, \dots, 9

f (x) = min {[x - 1], [x - 2], \dots, [x - 10]} = [x - 10] = k - 10

.

Thus

f (x) = [x] - 10

[0, 10)

\int_{0}^{10} f (x) d x = \sum_{k = 0}^{9} (k - 10) = - 10 - 9 - \dots - 1 = - 55

\int_{0}^{10} (f (x))^{2} d x = \sum_{k = 0}^{9} (k - 10)^{2} = 1 0^{2} + 9^{2} + \dots + 1^{2} = \frac{10 \cdot 11 \cdot 21}{6} = 385

\int_{0}^{10} ∣ f (x) ∣ d x = \sum_{k = 0}^{9} ∣ k - 10∣ = 10 + 9 + \dots + 1 = 55

.

Total:

- 55 + 385 + 55 = 385

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