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Question

If f(x) is monotonic differentiable function on [a,b], then baf(x)dx+f(b)f(a)f1(x)dx

A
bf(A)af(B)
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B
bf(B)af(A)
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C
f(A)+f(B)
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D
Cannot be found
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Solution

The correct option is B bf(B)af(A)
Consider I=baf(x)dx+f(b)f(a)f1(x)dx
I1 I2
take I2=f(b)f(a)f1(x)dx put x=f(t)
dx=f1(t)dt
=abf1(f(+)f1dt
=batf1(t)dt=tf(t)|ba=abf(t)dt
I2+I1=bf(b)af(a)
I=bf(b)af(a)


















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