Indefinite integrals

If $G(t)$ and $f(t)$ are such that $dG = fdt$, or equivalently $G'(t) = f(t)$, then $G$ is called an indefinite integral, or anti-derivative, or primitive of $f$.
We then write $G(t) = \int f(t)dt$, read as the indefinite integral of $f$.
So for example $t^3 - t^2 + 1$ is an indefinite integral of $3t^2 -2t$, since $d(t^3 - t^2 + 1) = (3t^2 - 2t)dt$.
Using the idea of an indefinite integral, we may rewrite FTCII as:

• $\int_a^b f(t) dt = G(b) - G(a)$, where $G(t) = \int f(t)dt$.

We then have a systematic way of writing out the calculation of an integral:

• If $f(t)$ has indefinite integral $G(t)$, we write:
  $$\int_a^b f(t) dt = [G(t)]_a^b = G(b) - G(a).$$

Examples:

$$\int_1^5 t^3 dt = [\frac{t^4}{4}]_1^5 = \frac{5^4}{4} - \frac{1}{4} = \frac{624}{4} = 156.$$

$$\int_1^5 \frac{1}{t^3} dt = [-\frac{1}{2t^2}]_1^5 = -\frac{1}{50} - (-\frac{1}{2}) = \frac{12}{25}.$$

$$\int_1^5 \frac{1}{t} dt = [\ln(\vert t\vert)]_1^5 = \ln(5) - \ln(1) = \ln(5).$$

$$\int_0^\frac{\pi}{2}\cos(t) dt = [\sin(t)]_0^\frac{\pi}{2} = \sin(\frac{\pi}{2}) - \sin(0) = 1 - 0 = 1.$$

$$\int_1^5e^{2t} dt = [\frac{1}{2}e^{2t}]_1^5 = \frac{1}{2}(e^{10} - e^2).$$