Integrals - First fundamental theorem of integral calculus

Find the derivative of $F(x) = \int_{2}^{x} \sin(t^2) \, dt$ with respect to $x$.

$F'(x) = \frac{d}{dx} \left[ \int_{2}^{x} \sin(t^2) \, dt \right] = \sin(x^2)$

According to the First Fundamental Theorem of Integral Calculus, $\frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x)$. Here, $f(t) = \sin(t^2)$, so we simply replace $t$ with $x$ in the integrand.

Find the value of $f(x)$ if $\int_{0}^{x} f(t) \, dt = x^2 + \cos(x) - 1$.

$\frac{d}{dx} \left[ \int_{0}^{x} f(t) \, dt \right] = \frac{d}{dx} (x^2 + \cos(x) - 1)$ $f(x) = 2x - \sin(x)$

To find $f(x)$ from the integral, we differentiate both sides with respect to $x$. By FTC 1, the derivative of the left side is $f(x)$. The derivative of the right side is obtained using standard differentiation rules.

Calculate the derivative of $G(x) = \int_{1}^{x^3} e^t \, dt$.

$G'(x) = e^{(x^3)} \cdot \frac{d}{dx}(x^3)$ $G'(x) = e^{x^3} \cdot 3x^2 = 3x^2 e^{x^3}$

When the upper limit is a function of $x$, say $g(x)$, we use the chain rule variation of FTC 1: $\frac{d}{dx} \int_{a}^{g(x)} f(t) \, dt = f(g(x)) \cdot g'(x)$. Here $g(x) = x^3$ and $f(t) = e^t$.