22. Riemann Sums, Integrals and the FTC

d. Integration Rules and Properties

1. Algebraic Operations

Since indefinite integrals are just general antiderivatives, all the rules for antiderivatives immediately become rules for indefinite integrals. In general, a derivative rule produces an indefinite integral rule: \[ \dfrac{d}{dx}[F(x)] =f(x) \qquad \text{becomes} \qquad \int f(x)\,dx=F(x)+C \]

So far, we have learned the derivative rules and corresponding integral rules shown in this table about algebraic operations and the one on the next page about special functions.

Derivative and Integral Rules for Algebraic Operations
Derivative Rule Integral Rule
Constant \(\dfrac{d}{dx}(c) =0\) Integrals are only determined up to a constant of integration.
Identity \(\dfrac{d}{dx}(x) =1\) \(\displaystyle \int 1\,dx=x+C\)
Power \(\dfrac{d}{dx}(x^n) =nx^{n-1}\) \(\displaystyle \int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C\)
Sum \(\dfrac{d}{dx}[f(x)+g(x)]=\dfrac{df}{dx}+\dfrac{dg}{dx}\) \(\displaystyle \int (f(x)+g(x))\,dx =\int f(x)\,dx+\int g(x)\,dx\)
Constant Multiple \(\dfrac{d}{dx}[cf(x)] =c\dfrac{df}{dx}\) \(\displaystyle \int cf(x)\,dx=c\int f(x)\,dx\)
Product \(\dfrac{d}{dx}[f(x) g(x)] =\dfrac{df}{dx}g(x)+f(x)\dfrac{dg}{dx}\)
Quotient \(\dfrac{d}{dx}\left[\dfrac{f(x)}{g(x)}\right] =\dfrac{g(x)\dfrac{df}{dx}-f(x)\dfrac{dg}{dx}}{[g(x)]^2}\)
Extended Power\(^\text{*}\) \(\dfrac{d}{dx}[g(x)^n]=n g(x)^{n-1}\dfrac{dg}{dx}\)
Chain \(\dfrac{d}{dx}[f(g(x))] =\left.\dfrac{df}{du}\right|_{u=g(x)}\dfrac{dg}{dx}\)

\(^\text{*}\) The Extended Power Rule is the special case of the Chain Rule for derivatives where \(f(u) =u^n\).

Noticeably missing from this table are integral rules corresponding to the Product, Quotient, Extended Power and Chain Rules for derivatives. In the next chapter, we will learn that the integral rule corresponding to the Chain Rule is called Integration by Substitution. (This includes the Extended Power Rule.) Then in Calculus 2, we will learn that the integral rule corresponding to the Product Rule is called Integration by Parts. To show/hide these additional rules in the table, click here. There is no nice integral rule corresponding to the Quotient Rule.

The following exercise is the same as on a previous page except solved using integrals and the Fundamental Theorem of Calculus.

Compute \(\displaystyle \int 12x^5-9x^2+4x+5\,dx\) and \(\displaystyle \int_1^2 12x^5-9x^2+4x+5\,dx\).

\(\displaystyle \int 12x^5-9x^2+4x+5\,dx=2x^6-3x^3+2x^2+5x+C\)
\(\displaystyle \int_1^2 12x^5-9x^2+4x+5\,dx=116\)

By the Sum, Constant Multiple and Power Rules, the indefinite integral is: \[\begin{aligned} \int 12x^5-9x^2+4x+5\,dx &=12\left(\dfrac{x^6}{6}\right)-9\left(\dfrac{x^3}{3}\right) +4\left(\dfrac{x^2}{2}\right)+5x+C \\ &=2x^6-3x^3+2x^2+5x+C \\ \end{aligned}\]

By the Fundamental Theorem of Calculus, the definite integral is: \[\begin{aligned} \int_1^2 &12x^5-9x^2+4x+5\,dx =\left[2x^6-3x^3+2x^2+5x\right]_1^2 \\ &=(2\cdot2^6-3\cdot2^3+2\cdot2^2+5\cdot2)-(2-3+2+5) \\ &=(128-24+8+10)-(6)=116 \end{aligned}\]

We check the indefinite integral by differentiating. If \(F=2x^6-3x^3+2x^2+5x\) then: \[ F'(x)=2\cdot6x^5-3\cdot3x^2+2\cdot2x+5=12x^5-9x^2+4x+5 \]

The Extended Power Rule is frequently used with trig functions.

Compute \(\displaystyle \int \cos^7(x)\sin(x)\,dx\) and \(\displaystyle \int_{\pi/6}^{\pi/3} \cos^7(x)\sin(x)\,dx\).

Apply the Extended Power Rule with \(g(x)=\cos(x)\).

\(\displaystyle \int \cos^7(x)\sin(x)\,dx=-\,\dfrac{\cos^8(x)}{8}+C\)
\(\displaystyle \int_{\pi/6}^{\pi/3} \cos^7(x)\sin(x)\,dx =\dfrac{5}{2^7}\)

We apply the Extended Power Rule with \(g(x)=\cos(x)\) and \(n=7\). Since \(\dfrac{d\cos(x)}{dx}=-\sin(x)\), we have \[\begin{aligned} \int \cos^7(x)\sin(x)\,dx &=-\,\dfrac{\cos^8(x)}{8}+C \\ \end{aligned}\] Be sure to check this by differentiating!

By the Fundamental Theorem of Calculus, the definite integral is: \[\begin{aligned} \int_{\pi/6}^{\pi/3} &\cos^7(x)\sin(x)\,dx =\left[-\,\dfrac{\cos^8(x)}{8}\right]_{\pi/6}^{\pi/3} \\ &=\dfrac{1}{8}\left[-\cos^8\left(\dfrac{\pi}{3}\right) +\cos^8\left(\dfrac{\pi}{6}\right)\right] \\ &=\dfrac{1}{8}\left[-\left(\dfrac{1}{2}\right)^8 +\left(\dfrac{\sqrt{3}}{2}\right)^8\right] \\ &=\dfrac{1}{8}\left[-\,\dfrac{1}{2^8}+\dfrac{3^4}{2^8}\right] =\dfrac{1}{8}\left[\dfrac{80}{2^8}\right] =\dfrac{5}{2^7} \end{aligned}\]

We check the indefinite integral by differentiating. If \(F=-\,\dfrac{\cos^8(x)}{8}\) then: \[ F'(x)=-\,\dfrac{1}{8}8\cos^7(x)[-\sin(x)]=\cos^7(x)\sin(x) \]

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