In this explainer, we will learn how to use indefinite integration to express a function given its rate of change.
At this point, we should be familiar with functions and their derivatives. However, sometimes we are not given an expression for a function, . Sometimes, we are only given an expression for the rate of change of that function, . When this happens, to find the original function, we would need to find antiderivatives, and to do that we will need to use integration.
When we first started learning how to integrate, we saw that indefinite integration can be used to find antiderivatives.
For example, we know that if then we can find the integral of this expression by using the power rule for integration:
The expression is called the most general antiderivative of our original function because it is an antiderivative for any value of .
To see this, we can differentiate with respect to : then we get our original function back.
However, this is only half of the story when it comes to integration. We can do exactly the same but this time we will start with
We can then follow the same line of working to see
In both of these equations, we differentiate a function to get ; in other words, . Integration and differentiation being reverse processes of one another is in fact a direct result of the fundamental theorem of calculus.
Therefore, if we start with an expression for the derivative of a function , we know that is an antiderivative of . So, we can try using indefinite integration to find an expression for . Of course, when we do this, we will calculate the most general antiderivative, which will contain an unknown constant, .
Remember, the derivative of any constant will be equal to 0, so can be any real number. This means that, to find the value of , we will need more information about our curve. Usually, this extra information is a point that lies on the curve; this is sometimes referred to as an initial condition or a boundary condition. We can then substitute the corresponding and values into the equation to find the value of .
Let’s see some examples of using this technique to find a function from its derivative and a point that lies on the curve.
Example 1: Finding the Equation of a Curve given the Expression of the Slope of Its Tangent and a Point on the Curve
Find the equation of the curve that passes through the point given that the gradient of the tangent to the curve is .
We are told that gradient of the tangent to a curve is equal to , but we also know that the gradient of a tangent line at a value of must be equal to (the rate of change of with respect to ).
We want to find an expression for our curve . Since the derivative of is , we must also have that the antiderivative of is equal to .
We can try to find this antiderivative by using indefinite integration:
We can evaluate this integral by using the power rule for integration:
So, we have found a general solution for our curve, which includes the constant of integration . To find the value of , we can use the fact that the curve passes through the point .
Therefore, our equation must hold when and :
Finally, we substitute this value of back into our equation of the curve to get
In our last example, we found the general solution . This will be an antiderivative for any value of . We can even sketch this curve for a few values of and consider what happens to the slope.
We can see that curves of this form are vertical translations of each other and that a vertical translation will not change the slope of the curve at any specific value of . For example, we can see that each of the three curves shown has the same slope when .
The curve is the only curve in this form which passes through the point ; hence, it is our specific solution.
In our next example, we will see that we cannot always directly use integration; we will often need to use some techniques to make the integration easier.
Example 2: Finding the Equation of a Curve given the Expression of the Slope of Its Tangent Using Integration and Trigonometric Identities
Find the equation of the curve given that the gradient of the tangent is and the curve passes through the origin.
First, we are told that the gradient of the tangent to a curve at a value of is given by . Remember that the gradient of a tangent line to a curve is also given by (the rate of change of with respect to ).
So, we are actually told in the question that and we want to find an equation for our curve.
We know that , so in other words we are looking for an antiderivative to . One way of doing this is to use indefinite integration:
We do not know how to directly integrate this function, so instead we are going to have to use the double-angle identity for cosine:
We rearrange this to see then we substitute to get
Now that we have rewritten our integrand in this form, we can find the indefinite integral of this expression:
This means we have now found a general solution for our curve up to the constant of integration . To find the value of , we need to recall that we are told that the curve passes through the origin.
Therefore, we can substitute and into our equation for the curve: which gives us
Substituting into the general equation of our curve means that we have just shown that the equation of the specific curve that passes through must be
Example 3: Finding the Value of a Function given the Expression of Its Slope by Using Indefinite Integration
Given that the slope at is and , determine .
The question gives us the slope of a curve. We are also told one output of our function at a given value of ; we need to use this to determine .
First, since we are given the slope of the curve representing our function , we can write this as .
We want to find , which is an antiderivative of . We can find this by using indefinite integration:
So, we must have that , for some value of . To find the value of we can use the fact that .
This gives us
Therefore, and we can substitute this into our equation for , giving us
Finally, the question wants us to find , so we just substitute into our function:
This gives us our final answer: .
So far, all of our examples have involved finding a function from its derivative; however, this is not the only thing we can do. Remember, derivatives can also be used to find the turning points of functions and the points where a curve changes concavity.
This means that when we are given the derivative of a function, we can use indefinite integration to help us find the equation of our curve. We can then use the given derivative in conjunction with the equation of the curve to find critical points.
Let’s see an example where we are given the slope of a curve and we want to find the coordinates of its local extrema.
Example 4: Finding the Local Maximum and Minimum Values of a Function given the Slope of Its Tangent and a Point on the Curve Involving Using Integration
Find the local maximum and minimum values of the curve that passes through the point where the gradient of the tangent is .
The first thing we need to notice is that, in the question, we are told that the gradient of the tangent to our curve is given by . This means that
We want to find the local maximum and minimum values of our curve; to do this, we need to remember that local extrema always occur at critical points in our function (where the derivative is equal to zero or it does not exist). In this case, the derivative we are given is a polynomial, which we know is well defined for all values of . Therefore, to find the critical points, we just need to find the values of for which the derivative is equal to 0:
So, the critical points of our curve will be when and . But we are not done; the question wants us to actually find the value of the local extrema, and to do this, we will need to find an expression for our curve.
Since our curve is an antiderivative of , we will find its equation by using indefinite integration:
This means that we must have that , for some value of . To find the value of , we need to use the fact that our curve passes through the point :
We can then substitute this value for back into the equation of our curve, giving us
We want to find the values of the local extrema, so we should start with evaluating our curve at the critical points.
We might be tempted to stop here; however, it is very important that we check to make sure these are in fact local extrema and not points of inflection, and which type of extrema these are. There are a few different options for doing this; for example, we could use the first or second derivative test. However, since our curve is a cubic with positive leading coefficient and two critical points, we can sketch our curve.
Given the shape of our curve, we can see that the lower value of occurs at and the greater value of occurs at . This gives us our final answer: the local maximum of our curve occurs at with a value of 15 and the local minimum of our curve occurs at with a value of 7.
Finally, it is also possible that we might need to use indefinite integration multiple times to find our original function. A good example of this is when we are given second derivative of a function and asked to find the original function; we can integrate once to find the slope function and then integrate this again to find the original function. Let’s see an example of this.
Example 5: Finding the Equation of a Curve given Its Second Derivative and the Equation of Its Tangent at a Point by Using Integration
The second derivative of a function is and the equation of the tangent to its graph at is . Find the equation of the curve.
We are given the second derivative of a function and asked to find the equation of the curve represented by this function. To do this, we need to recall what the second derivative of a curve means:
The second derivative of a function can be found by differentiating the first derivative.
Another way of phrasing this is that is an antiderivative of , so we can try to find by using integration:
At this point, we have two options. Our first option would be to immediately integrate the general slope function we have just found. This would introduce another unknown into our equation. Our second option would be to find the value of by using the equation of the tangent line to our curve. In this case, it is easier to work with the tangent line because we already have this information.
To do this, remember that we have ; this tells us the slope of the tangent line to our curve at . We are also told in the question that the tangent line to our curve at has the equation . We can find the slope of this line by writing the equation in slope–intercept form: which then tells us that
So, the tangent line when has a slope of 6. We can then substitute into the general equation we found for .
Since we also know the value of , we can use this to find our unknown:
This means we have shown that
We can now try to find an equation for our curve. We will do this by remembering that will be an antiderivative of :
So, we have just shown that the general solution for our curve is for some value of .
We can find the value of by using the fact that the curve passes through the point . Substituting these values into the equation of our curve, we get
Finally, we substitute this value of into the equation of our curve to get our final answer:
Let’s finish by recapping some of the points covered.
- Indefinite integration can be used to find the most general antiderivative of a function.
- If we are told that , then we know that
- We can find the value of the constant of integration, and hence the specific antiderivative, if we are given a point on the curve.
- When we are given the coordinates of a point that lies on the curve, this is referred to as an initial condition or a boundary condition.
- When working with higher-order derivatives, multiple integrals (and/or boundary conditions) may be needed to find the correct antiderivative.