(I2) Definite Integrals#

By the end of the lesson you will be able to:

  • evaluate a definite integral using the Fundamental Theorem of Calculus

Lecture Videos#

The Definite Integral#

Fundamental Theorem of Calculus (Part 2)

Suppose that \(f\) is a continuous function on an interval \([a,b]\) with an antiderivative \(F\). The definite integral of \(f\) from \(a\) to \(b\) can be calculated as:

\[ \int_a^{b} f(x)\, dx = F(b)-F(a) \]

Terminology#

The expression on the left side of this equation \(\int_a^b f(x) \, dx\) is the definite integral, and this theorem tells us how to calculate it.

  • The numbers \(a\) and \(b\) are called the limits of integration.

  • Specifically, \(a\) is called the lower limit and \(b\) is the upper limit.

How to Calculate a Definite Integral#

  1. Find the antiderivative of the function you’re integrating.

  2. Evaluate the antiderivative (plug the top number in first, the bottom number in second, and subtract.)

\[\begin{split} \begin{aligned} \int_a^{b} f(x)\, dx & \stackrel{(1)}{=} F(x) \bigg|_{x=a}^{x=b} \\ & \\ & \stackrel{(2)}{=} F(b)-F(a)\\ \end{aligned} \end{split}\]

Indefinite vs Definite Integrals#

Indefinite Integral
\[ \int f(x) \; dx \]

results in a function (use \(+C\))

Definite Integral
\[ \int_a^{b} f(x)\, dx \]

results in a number (do not use \(+C\))

What about +C?#

When calculating a definite integral, we do not need to use the \(+C\) for our antiderivative. (Actually we’re choosing the particular antiderivative with \(C=0\).)

If we did include it in our calculations we would find that it gets canceled off every time:

\[\begin{split} \begin{aligned} \int_a^{b} f(x)\, dx & = \big(F(x) +C\big) \bigg|_{x=a}^{x=b} \\ & \\ & = \bigg( F(b)+C \bigg) - \bigg( F(a)+C \bigg)\\ & \\ & = F(b)+C - F(a) - C \\ & \\ & = F(b) - F(a) \\ \end{aligned} \end{split}\]

Example 1#

Compute the following definite integrals:

\[ \int_1^2 x^2 \, dx \qquad \int_{-1}^{1} e^t \, dt \qquad \int_{-1}^{3} \dfrac{1}{x^2} \, dx \]

Click through the tabs to see the solution for each integral.

\[\begin{split} \begin{aligned} \int_1^2 x^2 \, dx & = \tfrac{1}{3}x^3 \bigg|_{x=1}^{x=2} \\ & \\ & = \left( \tfrac{1}{3}\cdot 2^3 \right) - \left( \tfrac{1}{3}\cdot 1^3 \right) \\ & \\ & = \tfrac{8}{3} - \tfrac{1}{3} \\ & \\ & = \tfrac{7}{3} \\ \end{aligned} \end{split}\]
\[\begin{split} \begin{aligned} \int_{-1}^{1} e^t \, dt & = e^t \bigg|_{t=-1}^{t=1} \\ & \\ & = \left( e^1 \right) - \left( e^{-1} \right) \\ & \\ & = e - e^{-1} \\ \end{aligned} \end{split}\]

Before we calculate a definite integral we do need to check whether the function we are integrating is continuous over the given interval.

Here our function is \(f(x)=\dfrac{1}{x^2}\) and the interval is \([-1,3]\). Is this function continuous over this entire interval? No, it is discontinuous at \(x=0\).

Because of this continuity, we are not able to calculate this integral using the Fundamental Theorem of Calculus.

Properties of the Definite Integral#

All of the following properties are related to the function we are integrating and are really inherited from the properties for the indefinite integral.

Sum / Difference Rule

\[ \int_a^{b} \big( f(x) \pm g(x) \big) \, dx = \int_a^{b} f(x)\, dx \pm \int_a^{b} g(x)\, dx\]

Constant-Multiple Rule

\[ \int_a^{b} k\cdot f(x) \, dx = k \cdot \int_a^{b} f(x)\, dx \]

Example 2#

Compute the following definite integrals:

\[ \int_0^{\pi/2} \bigg(3 \cos t + 2 \bigg) \, dt \qquad \int_0^{2} \bigg(2x^3 - 6x + \dfrac{3}{x^2+1} \bigg) \, dx \]
\[ \int_{1}^{3} \dfrac{3x-2x^3}{4x^4} \, dx \]

Click through the tabs to see the solution for each integral.

\[\begin{split} \begin{aligned} \int_0^{\pi/2} \bigg(3 \cos t + 2 \bigg) \, dt & = \big( 3\sin t + 2t\big)\; \bigg|_{t=0}^{t=\pi/2} \\ & \\ & = \left( 3\sin \tfrac{\pi}{2} + 2\cdot \tfrac{\pi}{2} \right) - \left( 3\sin 0 + 2\cdot 0 \right) \\ & \\ & = \left( 3\cdot 1 + 2\cdot \tfrac{\pi}{2} \right) - \left( 3\cdot 0 + 2\cdot 0 \right) \\ & \\ & = 3+\pi \\ \end{aligned} \end{split}\]
\[\begin{split} \begin{aligned} \int_1^{3} \bigg(2x^3 - 6x + \dfrac{3}{x^2+1} \bigg) \, dx & = \left(2\cdot\tfrac{1}{4}x^4 - 6\cdot \tfrac{1}{2} x^2 + 3 \arctan x\right)\; \bigg|_{x=0}^{x=2} \\ & \\ & = \left(\tfrac{1}{2}x^4 - 3 x^2 + 3 \arctan x\right)\; \bigg|_{x=0}^{x=2} \\ & \\ & = \left(\tfrac{1}{2}\cdot 2^4 - 3 \cdot 2^2 + 3 \arctan 2\right) - \left(\tfrac{1}{2}\cdot 0^4 - 3 \cdot 0^2 + 3 \arctan 0\right) \\ & \\ & = \left(8 - 12 + 3 \arctan 2\right) - \left(0 - 0 + 3 \cdot 0\right) \\ & \\ & = - 4 + 3 \arctan 2\\ \end{aligned} \end{split}\]

The function we are integrating here is a quotient, so we need to first perform the division. (We’re not able to split up the integration into the top and bottom of the fraction.)

\[\begin{split} \begin{aligned} \int_{1}^{3} & \dfrac{3x-2x^3}{4x^4} \, dx \\ & \\ & = \int_{1}^{3} \left( \dfrac{3x}{4x^4} - \dfrac{2x^3}{4x^4} \right) \, dx\\ & \\ & = \int_{1}^{3} \left( \dfrac{3}{4x^3} - \dfrac{2}{4x} \right) \, dx\\ & \\ & = \int_{1}^{3} \left( \dfrac{3}{4}\cdot\dfrac{1}{x^3} - \dfrac{1}{2}\cdot \dfrac{1}{x} \right) \, dx\\ & \\ & = \int_{1}^{3} \left( \dfrac{3}{4}\cdot x^{-3} - \dfrac{1}{2}\cdot \dfrac{1}{x} \right) \, dx\\ \end{aligned} \end{split}\]

We have simplified the function down to the point where it is now the sum of two constant-multiple power functions. So it’s something we can integrate now:

\[\begin{split} \begin{aligned} & = \left( \dfrac{3}{4}\cdot \dfrac{1}{-2} x^{-2} - \dfrac{1}{2}\cdot \ln |x| \right)\; \bigg|_{x=1}^{x=3} \\ & \\ & = \left( -\dfrac{3}{8}\cdot \dfrac{1}{x^2} - \dfrac{1}{2}\cdot \ln |x| \right)\; \bigg|_{x=1}^{x=3} \\ & \\ & = \left( -\dfrac{3}{8}\cdot \dfrac{1}{3^2} - \dfrac{1}{2}\cdot \ln |3| \right) - \left( -\dfrac{3}{8}\cdot \dfrac{1}{1^2} - \dfrac{1}{2}\cdot \ln |1| \right)\\ & \\ & = \left( -\dfrac{1}{24} - \dfrac{1}{2}\cdot \ln 3 \right) - \left( -\dfrac{3}{8} - \dfrac{1}{2}\cdot 0 \right)\\ & \\ & = -\dfrac{1}{24} - \dfrac{1}{2}\cdot \ln 3 + \dfrac{9}{24} \\ & \\ & = \dfrac{1}{3} - \dfrac{1}{2}\cdot \ln 3 \\ \end{aligned} \end{split}\]

More Properties of the Definite Integral#

All of the following properties are related to the limits of integration.

Switch Limits of Integration

\[ \int_a^{b} f(x) \, dx = - \int_b^{a} f(x)\, dx \]

Same Limits of Integration

\[ \int_a^{a} f(x) \, dx = 0 \]

Matching Limits of Integration

\[ \int_a^{m} f(x) \, dx + \int_m^{b} f(x) \, dx= \int_a^{b} f(x) \, dx \]

While the first two properties are fairly straight-forward, it might be helpful to look at an example for the third property:

Example 3#

Calculate the integral \(\displaystyle \int_0^5 f(x)\; dx\), given the following information:

\[ \int_0^6 f(x) \; dx =8 \qquad \int_5^6 f(x) \; dx =7 \]