All functions are continuous and s is twice differentiable Find the following:
- $\displaystyle\frac{d}{\mathrm dt}\int_t^{t^2} v(w)\,\mathrm dw$
- $\displaystyle\int_t^{t^2} v(w)\,\mathrm dw$
- $\displaystyle\frac{d^2}{\mathrm dt^2}\int_t^{t^2} v(w)\,\mathrm dw$
Set
$$F(x) = \int_0^xv(y)dy$$ then
$$\int_t^{t^2} v(y)dy = F(t^2)-F(t)\implies \frac{d}{dt}\int_t^{t^2} v(y)dy = 2t F'(t^2)-F'(t) = 2tv(t^2) -v(t)$$