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Question

If ϕ(x)=x1/xsin(t2)dt, then find the value of ϕ(1).

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Solution

ϕ(x)=x1xsin(t2)dt
using leibitz rule we, get
ϕ(x)=sin((x)2)ddxxsin((1x)2)ddx1x
ϕ(x)=sinx.12x1/2sin(1x2)(1x2)
ϕ(1)=sin12+sin11
ϕ(1)=32sin1

1078767_1094671_ans_20ac733e36cd46b6b0ccaf81dc7ba490.png

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