Integral Proving

Question

1.Find integral (from 0 to pi) [g"(t)]dt given that g(t) = integral (from 1 to sin(t)) [sqrtx^2+1]dx

Answer 1

g(t) = integral (from 1 to sint)[sqrt(x^2+1]dx

By the Fundamental Theorem of Calculus (FTC), g'(t) = sqrt[(sint)^2+1](sint)'

So, g'(t) = (cost)sqrt[(sint)^2 + 1]

Therefore, integral (0 to pi) g"(t) dt

= g'(t) (from 0 to pi)

= g'(pi) - g'(0)

= -1 - 1 = -2