미적분의 기본정리

(a) When $x=3$,

$$\begin{aligned} F(3)&=\displaystyle\int_0^3 3^3 \sin(t)\ \mathrm{d}t=27\int_0^3 \sin t \ \mathrm{d}t \\ \text{Using the Fundamental Theorem of Calculus Part 2,} \\ &=27\left[-\cos t\right]_{t=0}^{t=3} =27\left[-\cos 3 - (-\cos 0)\right]\\ &=27(1-\cos 3 ) \end{aligned}$$ (b) Since the integration is with respect to $t$, the $x^3$ term can be moved outside the integral. That is: for the purposes of the integral, $x^3$ is a constant (although for the purposes of the derivative, it certainly is not).

$$\begin{aligned} F(x)&=\displaystyle\int_0^x x^3 \sin(t)\ \mathrm{d}t = x^3 \int_0^x \sin(t)\ \mathrm{d}t \\ \text{Using the product rule and the Fundamental Theorem of Calculus Part 1,} \\ F'(x)&=x^3\cdot \sin(x) + 3x^2 \int_0^x \sin(t)\ \mathrm{d}t\\ &=x^3\sin(x)+3x^2\left[-\cos(t)\right]_{t=0}^{t=x}\\ &=x^3\sin(x)+3x^2[-\cos(x)-(-\cos(0))]\\ &=x^3\sin (x) + 3x^2[1-\cos (x)] \end{aligned}$$

Remark: Since $x$ and $t$ play different roles in our problem, it's crucial that they have different names. This is one reason why we should avoid the common mistake of writing $\int_a^x f(x)\mathrm{d}x$ when we mean $\int_a^x f(t)\mathrm{d}t$.