Integration
Integration (or antidifferentiation) is the opposite of differentiation. The notation for integration looks something like this:
\[\int_{a}^{b} f(x) \, dx\]
a and b are the terminals. We will discuss this in greater depth in the “definite integral” section.
dx indicates that we are integrating with respect to x. (i.e dealing with x)
There are many different methods we can use to estimate the area under a graph. In methods we look at two - the left and right endpoint methods.
Left endpoint method
To find the approximation of the area under a graph from x = a to x = b
1 | Divide the region into rectangles of equal width (this width will be given in the question). Dividing the x-axis into n equal subintervals. |
2 | Make sure that the top left corner of each rectangle touches the graph. The top right corner may or may not touch the graph. |
3 | Sum the area of the rectangles |
Using the left endpoint method to find the area under the graph \(y = (\frac{1}{12}x^2 + 1)\text{ from }( x= 0)\text{ to }(x = 8)\) with rectangles of width 2. This underestimates the area.
Right endpoint method
To find the approximation of the area under a graph from x = a to x = b
1 | Divide the region into rectangles of equal width (this width will be given in the question). Dividing the x-axis inton equal subintervals. |
2 | Make sure that the top right corner of each rectangle touches the graph. The top left corner may or may not touch the graph. |
3 | Sum the area of the rectangles |
Using the right endpoint method to find the area under the graph \(y = (\frac{1}{12}x^2 + 1)\text{ from }( x= 0)\text{ to }(x = 8)\) with rectangles of width 2. This overestimates the area.
The smaller the width of the rectangles, the more accurate the estimation of the area will be.
\[ \int k \cdot f(x) \,dx = k \cdot \int f(x) \, dx \] | \[\int f(x) + g(x) \,dx = \int f(x) \,dx + \int g(x)\,dx \] |
\[\int f(x) - g(x) \,dx = \int f(x)\, dx - \int g(x)\,dx \] | \[\int_{a}^{c} f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx\] |
\[\int_{a}^{b} f(x) \,dx = -\int_{b}^{a} f(x)\, dx \] | \[\int_{a}^{a} f(x) \, dx = 0\] |
\[\int x^n \, dx = \frac{1}{n+1}x^{n+1} + c ,\, n \neq -1 \] | \[\int (ax+b)^n \, dx = \frac{1}{x(n+1)}(ax+b)^{n+1} + c ,\, n \neq -1\] |
\[\int e^{ax+b} \,dx = \frac{1}{a}e^{ax+b} + c\] | \[\int\frac{1}{ax+b}\,dx = \frac{1}{a} \log_{e}{\lvert ax+b \rvert} + c \] |
\[\int \sin(ax+b) \,dx = -\frac{1}{a} \cos(ax+b) + c\] | \[\int \cos(ax+b) \,dx = \frac{1}{a} \sin(ax+b) + c\] |
The indefinite integral is denoted: \[\int f’(x) \, dx = f(x) + c\] \[c \text{ is an arbitrary constant, meaning that it could be any real number.}\] The inclusion of the arbitrary constant when dealing with indefinite integrals is essential, as you are indicating that there are infinitely many solutions (by varying the constant) not simply one solution. To find the value of the arbitrary constant, more information must be given.
\[ \text{Example 8.1: Find the rule of the function that passes through the point}\\ \text{ (1,3) and has gradient with the equation } \frac{dy}{dx} = 3x^2 + 3\\ \text{ }\\ \begin{aligned} y &= \int \frac{dy}{dx} \,dx\\ &= \int 3x^2 + 3 \, dx\\ &= \frac{3x^3}{3} + 3x + c ... \boxed{1} \\ \end{aligned}\\ \text{Substituting } x=1 \text{ and } y = 3 \text{ into } \boxed{1}:\\ \begin{aligned} 3 &= \frac{3\cdot 1^3}{3} + 3\cdot 1 + c\\ &= 1 + 3 + c\\ -1 &= c\\ y &= x^3 + 3x -1\\ \end{aligned} \]
Note that integrating products and fractions of functions (with a product/quotient rule like technique) is beyond the scope of this course. Below we detail how you should tackle these sorts of questions.
\[ \text{Example 8.2: Evaluate } \int (x-2)(x-4) \, dx\\ \text{ } \\ \text{For all products of functions, simply expand everything out. }\\ \begin{aligned} \int (x-2)(x-4)\, dx &= \int (x^2-6x+8) \,dx\\ &= \frac{x^3}{3} - \frac{6x^2}{2} + 8x + c\\ &= \frac{x^3}{3} -3x^2 +8x + c\\ \end{aligned}\\ \text{ }\\ \text{ }\\ \text{Example 8.3: Evaluate } \int \frac{6x^3-2x^2+9}{3x^2}\,dx\\ \text{ } \\ \text{For all fractions of functions, split the fraction up.}\\ \begin{aligned} \int \frac{6x^3-2x^2+9}{3x^2}\,dx &= \int \frac{6x^3}{3x^2} - \frac{2x^2}{3x^2} + \frac{9}{3x^2}\,dx\\ &= \int 2x - \frac{2}{3} + 3x^{-2}\,dx\\ &= \frac{2x^2}{2} -\frac{2x}{3} +(-3x^{-1})+c\\ &= x^2 -\frac{2x}{3} - 3x^{-1} + c\\ \end{aligned}\\ \]
Definite integrals look very similar to indefinite integrals, however, they have terminals either side of the integral symbol: \[ \begin{aligned} \int_{a}^{b} f'(x) \, dx &= [f(x)]_{a}^{b} \\ &= f(b) - f(a) \end{aligned} \]
Note that the final answer of a definite integral is a value, not an expression. The arbitrary constant is no longer present. After integrating the function, place it in square brackets with the terminals to the right of the square brackets like above. After that substitute the upper number (by position, not value) into the function and subtract the lower (by position, not value) number substituted into the function.
\[ \text{Example 8.4: Evaluate } \int_{1}^{-5} 3x^2 + 3 \, dx\\ \text{ } \\ \begin{aligned} \int_{1}^{-5} 3x^2 + 3 \, dx\ &= [\frac{3x^3}{3} + 3x]_{1}^{-5}\\ &= [x^3 + 3x]_{1}^{5}\\ &= ((-5)^3 + 3\cdot (-5)) - (1^3+3\cdot 1)\\ &= -125 - 15 -1 - 3\\ &= -144\\ \end{aligned} \]
To find the area of a region between two curves from \(x=a\) and \(x=b\), where \(f(x) \leqslant g(x)\) for \(x \in [a,b]\), find the difference between the areas under the graph of \(y = f(x)\) and the graph \(y = g(x)\) over the interval \([a,b]\). That is, evaluate: \(\int_{a}^{b} f(x) - g(x) \,dx\). In other words. Every time the graphs cross, you will need to create a new integral. For each integral, the equation of the graph which is visually lower is subtracted from the equation of the graph higher up. The lower terminal will be the smaller value in your interval and the upper terminal will be the larger value in your interval.
It is imperative to graph all equations that you are dealing with before attempting to find the area so that you know which equation has the larger values and for what x-values.
Note that the x-axis has equation y = 0
Below we have 3 examples to illustrate all cases you will encounter.
\[ \text{Example 8.5: Find the area bounded by the curve } y = \sin(x) \text{ and the x-axis for } x \in [0,\pi].\\ \text{ } \\ \text{Looking at the graph, } \sin(x) \text{is always above the x-axis,}\\ \text{ so we will subtract the equation of the x-axis from } \sin(x)\\ \] \[ \begin{aligned} \text{Area } &= \int_{0}^{\pi} \sin(x) - 0 )\, dx\\ &= [- \cos(x) ]_{0}^{\pi}\\ &= -\cos(\pi) -(-\cos(0))\\ &= -(-1) - (-1)\\ &= 2 \text{ units}^2\\ \end{aligned}\\ \]
\[ \text{Example 8.6: Find the area between the curves } y = e^x+3 \text{ and } y = -1 \text{ from } x = 1 \text{ to } x = 4.\\ \text{ } \\ \text{Looking at the graph, } e^x+3 \text{is always above } y = -1 \text{, so we will subtract } -1 \text{ from } e^x+3 \\ \] \[ \begin{aligned} \text{Area } &= \int_{1}^{4} (e^x + 3 - (-1)) \, dx\\ &= \int_{1}^{4} (e^x + 4) \, dx\\ &= [e^x + 4x]_{1}^{4}\\ &= (e^4 + 4 \cdot 4) - (e^1 + 4\cdot 1)\\ &= e^4 - e + 12 \text{ units}^2\\ \end{aligned}\\ \]
\[ \text{Example 8.7: Find the area bounded by the curves } y = \sin(x) \text{ and } y = \sin(2x) \text{ for } x \in [0,\pi].\\ \text{Looking at the graph, the two graphs clearly cross at one point.}\\\text{ We will need to find the intersection point.}\\ \text{For the first segment the graph of } \sin(2x) \text{ is higher up} \\ \text{But, for the second segment, the graph of } \sin(x) \text{ is higher up}\\ \] \[ \begin{aligned} \sin(x) &= \sin(2x)\\ x &= \frac{\pi}{3} \text{ (Using CAS calculator)}\\ \end{aligned}\\ \text{For } x \in [0,\frac{\pi}{3}], \sin(2x) \leqslant \sin(x) \text{, but for } x \in [\frac{\pi}{3}, \pi] \sin(x) \leqslant \sin(2x).\\ \text{ So, we must setup two integrals to calculate the combined area.}\\ \begin{aligned} &= \int_{0}^{\frac{\pi}{3}} \sin(2x) - \sin(x) \,dx + \int_{\frac{\pi}{3}}^{\pi} \sin(x) - sin(2x) \,dx\\ &= [-\frac{\cos(2x)}{2} - (-\cos(x))]_{0}^{\frac{\pi}{3}} + [-\cos(x) - (-\frac{\cos(2x)}{2})]_{\frac{\pi}{3}}^{\pi}\\ &= [-\frac{\cos(2x)}{2} + \cos(x)]_{0}^{\frac{\pi}{3}} + [-\cos(x) + \frac{\cos(2x)}{2}]_{\frac{\pi}{3}}^{\pi}\\ &= -\frac{\cos(2 \cdot \frac{\pi}{3})}{2} + \cos(\frac{\pi}{3}) - (-\frac{\cos(2 \cdot 0)}{2} + \cos(0)) + (- \cos(\pi) + \frac{\cos(2 \cdot \pi)}{2}) - (-\cos(\frac{\pi}{3}) + \frac{\cos(2 \cdot \frac{\pi}{5})}{2})\\ &= -\frac{\cos(\frac{2\pi}{3})}{2} + \cos(\frac{\pi}{3}) - (-\frac{\cos(0)}{2} + \cos(0)) + (- \cos(\pi) + \frac{\cos(2\pi)}{2}) - (-\cos(\frac{\pi}{3}) + \frac{\cos(2 \cdot \frac{\pi}{3})}{2})\\ &= -\frac{-\frac{1}{2}}{2} + \frac{1}{2} + \frac{1}{2} -1 - (-1) + \frac{1}{2} + \frac{1}{2} - \frac{(-\frac{1}{2})}{2}\\ &= \frac{1}{4} + \frac{1}{2} + \frac{1}{2} -1 + 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}\\ &= 3 \text{ units}^2\\ \end{aligned}\\ \]
For any continuous function f over the interval [a,b], the signed area is calculated by evaluating: \[\int_{a}^{b} f(x) dx\] This is not the true area, it considers any area under the x-axis to be negative. The signed area is generally useless, but does have some application in kinematics which is discussed further down.
The average value (not to be confused with average rate of change!) of a function f over the interval [a,b] is equal to the height of a rectangle having the same area as the area under the graph over the same interval. That is, \[\frac{1}{b-a} \cdot \int_{a}^{b} f(x) dx\]
\[ \text{Example 8.8: Find the average value of the } f(x) \text{ over the interval } [2,5] , f(x) = x^2 + 1\\ \begin{aligned} \text{Average Value} &= \frac{1}{b-a} \cdot \int_{a}^{b} f(x)\, dx\\ &= \frac{1}{5-2} \cdot \int_{2}^{5} x^2 + 1\, dx\\ &= \frac{1}{3} [\frac{x^3}{3}+x]_{2}^{5} \\ &= \frac{1}{3} ((\frac{5^3}{3} + 5) - (\frac{2^3}{3} + 2)) \\ &= \frac{1}{3} ((\frac{125}{3} + 5) - (\frac{8}{3} + 2)) \\ &= \frac{1}{3} (\frac{125}{3} + 5 - \frac{8}{3} - 2) \\ &= \frac{1}{3} (\frac{117}{3} + 3) \\ &= \frac{117}{9} + 1 \\ &= 13 + 1 \\ &= 14\\ \end{aligned}\\ \]
In kinematics you will be given an equation for the displacement, velocity or acceleration of an object possibly with some more information. Below we detail how you should calculate different quantities. But, first let’s define some key terms
Term | Definition |
Position | An object’s location relative to a selected point. |
Displacement | The direct distance an object is from a selected point |
Distance | How far an object moves in total |
Velocity | The change in position relative to time. |
Speed | The change in distance relative to time |
Acceleration | The change in velocity relative to time |
Let x(t) be the expression representing the position of an object
Let v(t) be the expression representing the velocity of an object
Let a(t) be the expression representing the acceleration of an object
To get between position, velocity and acceleration we simply use differentiation or integration.
\[
\begin{aligned}
x’(t) &= v(t) \\
v’(t) &= a(t) \\
\int a(t) \, dt &= v(t) \\
\int v(t) \, dt &= x(t) \\
\end{aligned}
\]
Below we have assumed that you are able to obtain all of the equations for position, velocity and acceleration of an object. Note that, when integrating from acceleration to velocity for example, you will need to have extra information to find the exact equation of velocity due to the introduction of the arbitrary constant following the indefinite integral calculation.
Term | Standard Question | Calculation |
Position | Find the position of an object at time t = u | \[x(u)\] |
Displacement | Find the displacement of an object between time t = u and t = w | \[ x(w) - x(u) \\ \text{It is also the signed area of the graph v(t)} \\ \int_{u}^{w} v(t) \, dt \] |
Distance | Find the distance an object travels between t = u and t = w | The area bounded by the graph of v(t) and the x-axis. \[\int_{u}^{w} \lvert v(t) \rvert \, dt\] |
Velocity | Find the velocity of an object at t = u | \[v(u)\] |
Average velocity | Find the average velocity of an object between t = u and t = w | \[\frac{x(w)-x(u)}{w-u}\] |
Speed | Find the speed of an object at t = u | \[\lvert v(u) \rvert \] |
Average speed | Find the average speed of an object between t = u and t = w | Total distance travelled in the interval divided by the length of the time interval. \[\frac{\int_{u}^{w} \lvert v(t) \rvert \, dt}{w-u}\] |
Acceleration | Find the acceleration of an object at t = u | \[a(u)\] |
Deceleration | Find how fast the car is decelerating at t = u | Same as acceleration. However, deceleration is positive. That is, if acceleration is -8, then deceleration is 8. |
Average acceleration | Find the average acceleration of an object between t = u and t = w | \[\frac{v(w) - v(u)}{w-u}\] |
Common, but one of the harder integration questions. An example is given below. \[ \text{Example 8.9: If }f(x)=x\cdot \cos(3x),\text{ then }f^\prime(x)=\cos(3x)-3x\cdot \sin(3x)\\ \text{Use this fact to find an antiderivative of }x\cdot \sin(3x)\\ \begin{aligned} \text{ } \\ \text{} \int f^\prime(x)\,dx &= f(x) + c\\ \int \cos(3x)- 3x\cdot \sin(3x))\,dx &= x\cdot \cos(3x) + c\\ \int \cos(3x)\, dx - \int 3x\cdot \sin(3x))\,dx &= x\cdot \cos(3x) + c\\ \int \cos(3x)\, dx - 3\int x\cdot \sin(3x))\,dx &= x\cdot \cos(3x) + c\\ \int (x\cdot \sin(3x))\,dx&=\frac{1}{3}(\int \cos(3x)\,dx - x\cdot \cos(3x)) + c\\ &=\frac{1}{3}\cdot (\frac{1}{3}\cdot \sin(3x) - \cdot x\cdot \cos(3x)) + c\\ &=\frac{1}{3}\cdot \frac{1}{3}\cdot \sin(3x) - \frac{1}{3}\cdot x\cdot \cos(3x) + c\\ &=\frac{\sin(3x)}{9} - \frac{x\cdot \cos(3x)}{3} + c\\ \end{aligned}\\ \text{An antiderivative is } \frac{\sin(3x)}{9} - \frac{x\cdot \cos(3x)}{3} \, ,(c = 0) \\ \]