Integrating Constants
This page follows the JoeCMath video The Simplest Integral You’ll Ever Solve. The lesson works through constant integrals first without bounds, then with bounds, so the main question is whether the answer stays as an antiderivative or becomes one number.
Main Idea
When the integrand is a constant $k$, the antiderivative is linear:
\[\int k\,dx=kx+C.\]If bounds are included, the constant gets multiplied by the length of the interval:
\[\int_a^b k\,dx=k(b-a).\]The video uses $5$ to show both forms before moving to quick examples with $7$, $\pi$, and $125$.
Watch this section: introduction at 0:00.
Two Forms For The Same Constant
At the start, the video separates the two forms students can see when integrating a constant.
Without bounds, the integral is indefinite:
\[\int 5\,dx.\]With bounds, the integral is definite:
\[\int_1^6 5\,dx.\]The constant being integrated is still $5$ in both cases. The difference is what the notation asks for. The indefinite version asks for an antiderivative. The definite version asks for the accumulated value from the lower bound to the upper bound.
Watch this section: definite and indefinite forms at 0:05.
Indefinite Example
The first full example is
\[\int 5\,dx.\]The video rewrites the constant as $5\cdot 1$, then replaces $1$ with $x^0$ because $x^0=1$:
\[\int 5\,dx =\int 5\cdot 1\,dx =\int 5x^0\,dx.\]Now the integration power rule can be used:
\[\int 5x^0\,dx =5\cdot\frac{x^{0+1}}{0+1}+C.\]After simplifying,
\[\int 5\,dx=5x+C.\]The $+C$ appears because this example has no bounds.
Watch this section: first example at 0:15.
Definite Example
The second example keeps the same constant but adds bounds:
\[\int_1^6 5\,dx.\]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:
\[5(6)-5(1)=5(6-1)=25.\]That is why a definite integral of a constant can be read as the constant times the interval length.
Watch this section: second example at 1:38.
Formula Review
The review section names the two formulas side by side. For an indefinite integral of a constant,
\[\int k\,du=ku+C.\]The variable in the answer matches the differential. If the integral ends in $dx$, use $x$. If it ends in $dy$, use $y$.
For a definite integral of a constant,
\[\int_a^b k\,du=k(b-a).\]There is no $+C$ in the final definite answer because the bounds turn the result into one number.
Watch this section: formula review at 2:34.
Rapid Fire Examples
The last worked section applies the formulas quickly.
For the indefinite integral
\[\int 7\,dx,\]the constant is $7$ and the differential is $dx$, so
\[\int 7\,dx=7x+C.\]For the definite integral
\[\int_3^{10}\pi\,dx,\]the constant is $\pi$ and the interval length is $10-3$, so
\[\int_3^{10}\pi\,dx=\pi(10-3)=7\pi.\]For the final indefinite example,
\[\int 125\,dy=125y+C.\]The answer uses $y$ because the integral is with respect to $y$.
Watch this section: rapid fire examples at 2:55.
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.