To determine

To find:

g''(π/6)

Answer

154

Explanation

1) Concept:

Fundamental Theorem of Calculus:

If f is a continuous function on [a, b], then the function g defined by

gx=∫axf(t) dt

is continuous on [a, b]  and differentiable on  (a, b), and g'x=f(x)

That is,ddx∫axf(t) dt=f(x)

2) Given:

fx=∫0sin⁡x1+t2dt and gy=∫3yf(x)dx

3) Calculation:

Here

gy=∫3yf(x) dx

By using Fundamental Theorem of Calculus,

g'y=f(y) and so g''y=f'(y)

Now

fx=∫0sin⁡x1+t2dt

Replace the variable x with y to get,

fy=∫0sin⁡y1+t2dt

To find f'y differentiate f(y),

f'y=ddy∫0sin⁡y1+t2dt

Let sin⁡y=u then

dudy=cos⁡y So

ddy∫0sin⁡y1+t2dt=ddy∫0u1+t2dt

By chain rule,

ddy∫0u1+t2dt=ddu∫0u1+t2dtdudy

By fundamental theorem of calculus

ddu∫0u1+t2dt=1+u2 So

ddy∫0sin⁡y1+t2dt=1+u2·cos⁡y

Thus substituting value of u we have

f'y=1+sin2⁡y·cos⁡y

Given that, y=π/6 So

f'π/6=1+sin2⁡π/6·cos⁡π/6

=1+122·32

Simplifying,

=44+14·32

=54·32

=5·34·12

=152·12

=154

So, value of g''π6=154

Conclusion:

Therefore,

g''π6=154