Calculus / Unit 8: The Fundamental Theorem of Calculus
8.3 FTC Part 1: Derivatives of Integrals
Notes
FTC Part 1: if $f$ is continuous and $\displaystyle G(x) = \int_a^x f(t)\,dt$, then $G'(x) = f(x)$.
- Differentiating an integral with a variable upper limit gives back the integrand, evaluated at $x$.
- If the upper limit is a function $u(x)$, use the chain rule: $\displaystyle \frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x))\,u'(x)$.
- Swapping the limits adds a negative sign: $\displaystyle \frac{d}{dx}\int_x^5 t^2\,dt = -x^2$.
Practice
Find $\displaystyle \frac{d}{dx}\int_2^x t^3\,dt$.
FTC Part 1 gives the integrand at $x$.
Find $\displaystyle \frac{d}{dx}\int_x^5 t^2\,dt$.
Reversing the limits gives $\displaystyle -\int_5^x t^2\,dt$, whose derivative is $-x^2$.
Find $\displaystyle \frac{d}{dx}\int_0^{x^2} \cos t\,dt$.
Chain rule: $\cos(x^2)$ times the derivative of the upper limit, $2x$.
Ready for more? Take the Unit 8 test.