Calculating Integrals

Section 4.5 Calculating Integrals

Standards

Enables: Integrate polynomial, trig, and/or exponential functions

We learned properties for breaking down integration calculations (dividing the interval, sums, scalar multiples) and we have learned from the Fundamental Theorem of Calculus that we can use derivatives (in reverse) to calculate integrals. Here we will illustrate how we can put these together to calculate more integrals.

Subsection 4.5.1 Using the FTC

Example 4.5.1.

We want to calculate

\begin{equation*} \int_0^{\pi/2} \cos\theta \; d\theta \end{equation*}

According to the FTC Part 2 4.4.2 we can do this if we know a function \(G(\theta)\) such that \(G^\prime(\theta)=\cos\theta\text{.}\) Thankfully, we know that the derivative of \(G(\theta)=\sin\theta\) is \(\cos\theta\text{.}\) Thus by the FTC

\begin{align*} \int_0^{\pi/2} \cos\theta \; d\theta & = \sin(\pi/2)-\sin(0)\\ & = 1 - 0\\ & = 1 \end{align*}

Example 4.5.2.

For convenience we have a notation to help us break up the steps. Below is the same problem as above using the evaluation notation.

\begin{align*} \int_0^{\pi/2} \cos\theta \; d\theta & = \left. \sin(\theta) \right|_0^{\pi/2}\\ & = sin(\pi/2)-\sin(0)\\ & = 1 - 0\\ & = 1 \end{align*}

Example 4.5.3.

We want to calculate

\begin{equation*} \int_1^5 3x^2 \; dx \end{equation*}

According to the FTC Part 2 4.4.2 we can do this if we know a function \(G(x)\) such that \(G^\prime(x)=3x^2\text{.}\) Thankfully, we know that the derivative of \(G(x)=x^3\) is \(3x^2\text{.}\) Thus by the FTC

\begin{align*} \int_1^5 3x^2 \; dx & = \left. x^3 \right|_1^5\\ & = 5^3-1^3\\ & = 125 - 1\\ & = 124 \end{align*}

Checkpoint 4.5.4.

Calculate

\begin{equation*} \int_1^3 5x^4 \; dx\text{.} \end{equation*}

Subsection 4.5.2 Using Multiple Integral Properties

We know the following from previous sections.

FTC: \(\int_a^b f(x) \; dx = G(b)-G(a)\) where \(G^\prime(x)=f(x)\text{.}\)
\(\displaystyle \int_a^b f(x)+g(x) \; dx = \int_a^b f(x) \; dx + \int_a^b g(x) \; dx\)
\(\displaystyle \int_a^b k \cdot f(x) \; dx = k \int_a^b f(x) \; dx\)

We can put these together to calculate more integrals.

Example 4.5.5.

We want to calculate \(\int_0^{10} 15x^2-4x+1 \; dx\text{.}\) Because we do not know off the top of our heads a function for which this is the derivative, we will need to use the properties.

\begin{align*} \int_0^{10} 15x^2-4x+1 \; dx = & \\ \int_0^{10} 15x^2 \; dx + \int_0^{10} -4x \; dx + \int_0^{10} 1 \; dx = & \text{ sum property}\\ \int_0^{10} 5 \cdot 3x^2 \; dx + \int_0^{10} -2 \cdot 2x \; dx + \int_0^{10} 1 \; dx = & \text{ turned into known derivatives}\\ 5 \int_0^{10} 3x^2 \; dx -2 \int_0^{10} 2x \; dx + \int_0^{10} 1 \; dx = & \text{ scalar product property (twice)}\\ 5 \left( \left. x^3 \right|_0^{10} \right) -2 \left( \left. x^2 \right|_0^{10} \right) + \left. x \right|_0^{10} = & \text{ FTC}\\ 5(10^3-0^3)-2(10^2-0^2)+(10-0) = & 4810 \end{align*}

Example 4.5.6.

Sometimes the integrand is not quite the derivative of a function, but we can adjust a coefficient to make it work.

\begin{align*} \int_1^5 x^4 \; dx & =\\ \int_1^5 \frac{1}{5} \cdot 5 x^4 \; dx & = \text{ insert needed coefficient}\\ \frac{1}{5} \int_1^5 5 x^4 \; dx & = \text{ scalar product property}\\ \frac{1}{5} \left. x^5 \right|_1^5 & = \text{ FTC}\\ \frac{1}{5} 5^5-1^5 & = 3124. \end{align*}

This technique if frequently used.

Checkpoint 4.5.7.

Calculate the following integral.

\begin{equation*} \int_0^1 2x^2-3x+6 \; dx \end{equation*}

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