Integral Calculus-2
Integral Calculus-2
Allen Star Batch
Grade 12

Question:

The value of $\int_1^8 x\sin[x^2 - \pi] dx$, where $[.]$ denotes the greatest integer function is:
$\sum_{r=1}^6 r\sin r$
$\sum_{r=1}^6 (-1)^r r\sin r$
$\sum_{r=1}^6 r^2 \sin r$
None of these

Step-by-Step Solution

Key Concept: The greatest integer function [x² - π] creates piecewise constant intervals on [1,8]. Since π ≈ 3.14, we have x² - π ranging from -2.14 to 61.86, so [x² - π] takes integer values 0,1,2,...,8 on disjoint intervals. The integral splits as Σ∫ r·x·sin(r)dx over intervals where [x² - π] = r, but evaluating these requires solving x² = r + π for bounds.
The integral $\int_{-π}^π x\sin(x^2-π)dx = 2\int_0^π x\sin(x^2-π)dx$ by symmetry considerations. Since $-π ≤ x^2 - π ≤ π^2 - π ≤ π^2 - 7$ and $-π ≤ x^2 - π$ for all $x$ in the domain, the last integral equals $2\int_0^π x\sin(x^2)dx$ by periodicity of sine.
Correct Answer: 4

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