Integral Calculus

Question

Find ∫₀^pi g"(t)dt given that g(t)=∫₁^sin(t)[√x²+1] dx

Answer 1

If g(t) = integral (from 1 to sint)[sqrt(x^2+1)]dx, then, by the Fundamental Theorem of Calculus,

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

So, integral (from 0 to pi)[g"(t)]dt = g'(t) (evaluated from 0 to pi)

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

= -1 - 1 = -2