Q.
If for every integer n,∫nn+1f(x)dx=n2, then the value of ∫−24f(x)dx is
2026
188
Rajasthan PETRajasthan PET 2012
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Solution:
Given, ∫nn+1f(x)dx=n2
On putting n=−2,−1,0,1,2,3, we get ∫−2−1f(x)dx=4∫−10f(x)dx=1 ∫01f(x)dx=0∫12f(x)dx=1 ∫23f(x)dx=4∫34f(x)dx=9 ∴∫−24f(x)dx=4+1+0+1+4+9=19