Indefinite Integration 1 Question 49

50. Let f(x)=192x32+sin4πx for all xR with f12=0. If m1/21f(x)dxM, then the possible values of m and M are

(2015 Adv.)

(a) m=13,M=24

(b) m=14,M=12

(c) m=11,M=0

(d) m=1,M=12

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Solution:

  1. Here, f(x)=192x32+sin4πx192x33f(x)192x32

On integrating between the limits 12 to x, we get

1/2x192x33dx1/2xf(x)dx1/2x192x32dx19212x4116f(x)f(0)24x43216x41f(x)24x432

Again integrating between the limits 12 to 1 , we get

1/21(16x41)dx1/21f(x)dx1/2124x432dx16x55x1/21f(x)dx24x5532x1/21115+251/21f(x)dx3310+6102.61/21f(x)dx3.9

) None of the option is correct.



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