Integrating Constants
Simple overview for finding the integral of a contstant.
Main Idea
When you are integrating a constant $k$, the integral $kx+C$ or $[kx]^b_a$.
\[\int k dx=kx+C\]If the integral is bounded, the constant gets multiplied by the length of the interval ($F(b)-F(a)=k(b-a)$)
\[\int_a^b k\,dx=k(b-a)\]Watch this section: introduction at 0:00.
Indefinite Example (Where does the rule comes from?)
Consider the following indefinite integral
\[\int 5 dx\]We rewrite $5$ as the product of $5$ and a power of our variable we are integrating ($x^0$ in this case).
\[\int 5 dx =\int 5\cdot 1 dx =\int 5x^0 dx\]Now the integration power rule can be used:
\[\int 5x^0\,dx =5\cdot\frac{x^{0+1}}{0+1}+C\]After simplifying,
\[\int 5 dx=5x+C\]Now, instead of having to introduce this strange form of $1$ over and over again we can use the rule to just straight to the answer.
\[\int k dx=kx+C\]Watch this section: first example at 0:15.
Definite Example
What happens when our integral is bounded?
\[\int_1^6 5 dx\]The antiderivative of $5$ is $5x$, so the bounded integral is evaluated from $1$ to $6$
\[\int_1^6 5 dx=[5x]_1^6\]Substitute the top bound first, then subtract the bottom bound
\[[5x]_1^6=5(6)-5(1)=30-5=25\]The video also factors the $5$ to show the shortcut hiding in the arithmetic:
\[5(6)-5(1)=5(6-1)=25.\]Watch this section: second example at 1:38.
Timestamp Guide
Related Calculus Work
The video turns a constant into $x^0$ so the integration power rule can be used.
For more on why indefinite integrals include $+C$, review the constant of integration notes.
For the broader contrast between answers with and without bounds, use the definite vs. indefinite integrals companion or continue through the full JoeCMath integration playlist.