Lesson Explainer: Applications of Indefinite Integration Mathematics

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, 𝑓(π‘₯). Sometimes, we are only given an expression for the rate of change of that function, 𝑓′(π‘₯). 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 𝑦=π‘₯, then we can find the integral of this expression by using the power rule for integration: ο„Έπ‘₯π‘₯=π‘₯3+.dC

The expression π‘₯3+C is called the most general antiderivative of our original function π‘₯ because it is an antiderivative for any value of C.

To see this, we can differentiate π‘₯3+C with respect to π‘₯: ddCπ‘₯ο€Ύπ‘₯3+=π‘₯, 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 dd𝑦π‘₯=π‘₯.

We can then follow the same line of working to see ddCπ‘₯ο€Ύπ‘₯3+=π‘₯.

In both of these equations, we differentiate a function to get π‘₯; in other words, 𝑦=π‘₯3+C. 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 𝑓′(π‘₯), we know that 𝑓(π‘₯) is an antiderivative of 𝑓′(π‘₯). So, we can try using indefinite integration to find an expression for 𝑓(π‘₯). Of course, when we do this, we will calculate the most general antiderivative, which will contain an unknown constant, C.

Remember, the derivative of any constant will be equal to 0, so C can be any real number. This means that, to find the value of C, 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 π‘₯ and 𝑦 values into the equation to find the value of C.

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 (βˆ’2,1) given that the gradient of the tangent to the curve is βˆ’11π‘₯.

Answer

We are told that gradient of the tangent to a curve is equal to βˆ’11π‘₯, but we also know that the gradient of a tangent line at a value of π‘₯ must be equal to dd𝑦π‘₯(the rate of change of 𝑦 with respect to π‘₯).

Therefore, dd𝑦π‘₯=βˆ’11π‘₯.

We want to find an expression for our curve 𝑦. Since the derivative of 𝑦 is βˆ’11π‘₯, we must also have that the antiderivative of βˆ’11π‘₯ is equal to 𝑦.

We can try to find this antiderivative by using indefinite integration: 𝑦=ο„Έβˆ’11π‘₯π‘₯.d

We can evaluate this integral by using the power rule for integration: 𝑦=ο„Έβˆ’11π‘₯π‘₯=βˆ’113π‘₯+.dC

So, we have found a general solution for our curve, which includes the constant of integration C. To find the value of C, we can use the fact that the curve passes through the point (βˆ’2,1).

Therefore, our equation must hold when π‘₯=βˆ’2 and 𝑦=1: 1=βˆ’113(βˆ’2)+1=883+=βˆ’853.CCC

Finally, we substitute this value of C back into our equation of the curve to get 𝑦=βˆ’113π‘₯βˆ’853.

In our last example, we found the general solution 𝑦=βˆ’113π‘₯+C. This will be an antiderivative for any value of C. We can even sketch this curve for a few values of C 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 π‘₯. For example, we can see that each of the three curves shown has the same slope when π‘₯=0.

The curve 𝑦=βˆ’113π‘₯βˆ’853 is the only curve in this form which passes through the point (βˆ’2,1); 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 5ο€»π‘₯2sin and the curve passes through the origin.

Answer

First, we are told that the gradient of the tangent to a curve at a value of π‘₯ is given by 5ο€»π‘₯2sin. Remember that the gradient of a tangent line to a curve is also given by dd𝑦π‘₯ (the rate of change of 𝑦 with respect to π‘₯).

So, we are actually told in the question that ddsin𝑦π‘₯=5ο€»π‘₯2, and we want to find an equation 𝑦=𝑓(π‘₯) for our curve.

We know that 𝑓′(π‘₯)=5ο€»π‘₯2sin, so in other words we are looking for an antiderivative to 5ο€»π‘₯2sin. One way of doing this is to use indefinite integration: 𝑦=ο„Έ5ο€»π‘₯2π‘₯.sind

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: cossin(2πœƒ)≑1βˆ’2πœƒ.

We rearrange this to see sincosοŠ¨πœƒβ‰‘1βˆ’(2πœƒ)2, then we substitute πœƒ=π‘₯2 to get sincosοŠ¨ο€»π‘₯2≑1βˆ’(π‘₯)2.

Now that we have rewritten our integrand in this form, we can find the indefinite integral of this expression: 𝑦=ο„Έ5ο€»π‘₯2π‘₯=ο„Έ5ο€½1βˆ’(π‘₯)2π‘₯=5ο„Έ1βˆ’(π‘₯)2π‘₯=5ο€½π‘₯2βˆ’(π‘₯)2+=5π‘₯2βˆ’5(π‘₯)2+.sindcosdcosdsinCsinC

This means we have now found a general solution for our curve up to the constant of integration C. To find the value of C, we need to recall that we are told that the curve passes through the origin.

Therefore, we can substitute π‘₯=0 and 𝑦=0 into our equation for the curve: 0=5(0)2βˆ’5(0)2+,sinC which gives us C=0.

Substituting C=0 into the general equation of our curve means that we have just shown that the equation of the specific curve that passes through (0,0) must be 𝑦=52π‘₯βˆ’5(π‘₯)2.sin

Example 3: Finding the Value of a Function given the Expression of Its Slope by Using Indefinite Integration

Given that the slope at (π‘₯,𝑦) is 3π‘’οŠ©ο— and 𝑓(0)=βˆ’3, determine 𝑓(βˆ’3).

Answer

The question gives us the slope of a curve. We are also told one output of our function at a given value of π‘₯; we need to use this to determine 𝑓(βˆ’3).

First, since we are given the slope of the curve representing our function 𝑓, we can write this as 𝑓′(π‘₯)=3π‘’οŠ©ο—.

We want to find 𝑓(π‘₯), which is an antiderivative of 3π‘’οŠ©ο—. We can find this by using indefinite integration: 𝑓(π‘₯)=ο„Έ3𝑒π‘₯=𝑒+.οŠ©ο—οŠ©ο—dC

So, we must have that 𝑓(π‘₯)=𝑒+οŠ©ο—C, for some value of C. To find the value of C we can use the fact that 𝑓(0)=βˆ’3.

This gives us βˆ’3=𝑓(0)=𝑒+=1+.οŠ©ο—CC

Therefore, C=βˆ’4, and we can substitute this into our equation for 𝑓(π‘₯), giving us 𝑓(π‘₯)=π‘’βˆ’4.οŠ©ο—

Finally, the question wants us to find 𝑓(βˆ’3), so we just substitute π‘₯=3 into our function: 𝑓(βˆ’3)=π‘’βˆ’4=π‘’βˆ’4=βˆ’4+1𝑒.()

This gives us our final answer: 𝑓(βˆ’3)=βˆ’4+1π‘’οŠ―.

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 (βˆ’1,7) where the gradient of the tangent is 6ο€Ήπ‘₯+4π‘₯+3ο…οŠ¨.

Answer

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 6ο€Ήπ‘₯+4π‘₯+3ο…οŠ¨. This means that dd𝑦π‘₯=6ο€Ήπ‘₯+4π‘₯+3.

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 π‘₯. Therefore, to find the critical points, we just need to find the values of π‘₯ for which the derivative is equal to 0: 6ο€Ήπ‘₯+4π‘₯+3=0,π‘₯+4π‘₯+3=0,(π‘₯+3)(π‘₯+1)=0,π‘₯+3=0,π‘₯+1=0.

So, the critical points of our curve will be when π‘₯=βˆ’3 and π‘₯=βˆ’1. But we are not done; the question wants us to actually find the 𝑦 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 6ο€Ήπ‘₯+4π‘₯+3ο…οŠ¨, we will find its equation by using indefinite integration: 𝑦=ο„Έ6ο€Ήπ‘₯+4π‘₯+3π‘₯=6ο„Έπ‘₯+4π‘₯+3π‘₯=6ο€Ύπ‘₯3+4π‘₯2+3π‘₯+=2π‘₯+12π‘₯+18π‘₯+.ddCC

This means that we must have that 𝑦=2π‘₯+12π‘₯+18π‘₯+C, for some value of C. To find the value of C, we need to use the fact that our curve passes through the point (βˆ’1,7): 7=2(βˆ’1)+12(βˆ’1)+18(βˆ’1)+7=βˆ’2+12βˆ’18+=15.CCC

We can then substitute this value for C back into the equation of our curve, giving us 𝑦=2π‘₯+12π‘₯+18π‘₯+15.

We want to find the values of the local extrema, so we should start with evaluating our curve at the critical points.

π‘₯=βˆ’1: 𝑦=2(βˆ’1)+12(βˆ’1)+18(βˆ’1)+15=7.

π‘₯=βˆ’3: 𝑦=2(βˆ’3)3+12(βˆ’3)2+18(βˆ’3)+15=15.

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 𝑦 occurs at π‘₯=βˆ’1 and the greater value of 𝑦 occurs at π‘₯=βˆ’3. This gives us our final answer: the local maximum of our curve occurs at π‘₯=βˆ’3 with a value of 15 and the local minimum of our curve occurs at π‘₯=βˆ’1 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 6π‘₯ and the equation of the tangent to its graph at (βˆ’2,βˆ’4) is 6π‘₯βˆ’π‘¦+8=0. Find the equation of the curve.

Answer

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: ddddddοŠ¨οŠ¨π‘¦π‘₯=π‘₯𝑦π‘₯.

The second derivative of a function can be found by differentiating the first derivative.

Another way of phrasing this is that dd𝑦π‘₯ is an antiderivative of ddοŠ¨οŠ¨π‘¦π‘₯, so we can try to find dd𝑦π‘₯ by using integration: ddddddC𝑦π‘₯=𝑦π‘₯π‘₯=ο„Έ6π‘₯π‘₯=3π‘₯+.

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 C 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 ddC𝑦π‘₯=3π‘₯+; this tells us the slope of the tangent line to our curve at π‘₯. We are also told in the question that the tangent line to our curve at (βˆ’2,βˆ’4) has the equation 6π‘₯βˆ’π‘¦+8=0. We can find the slope of this line by writing the equation in slope–intercept form: 𝑦=6π‘₯+8, which then tells us that dd𝑦π‘₯|||=6.ο—οŠ²οŠ±οŠ¨

So, the tangent line when π‘₯=βˆ’2 has a slope of 6. We can then substitute π‘₯=βˆ’2 into the general equation we found for dd𝑦π‘₯.

Since we also know the value of dd𝑦π‘₯|||=6ο—οŠ²οŠ±οŠ¨, we can use this to find our unknown: 6=3(βˆ’2)+,=βˆ’6.CC

This means we have shown that dd𝑦π‘₯=3π‘₯βˆ’6.

We can now try to find an equation for our curve. We will do this by remembering that 𝑦 will be an antiderivative of dd𝑦π‘₯: 𝑦=𝑦π‘₯π‘₯=ο„Έο€Ή3π‘₯βˆ’6π‘₯=π‘₯βˆ’6π‘₯+𝐷.dddd

So, we have just shown that the general solution for our curve is 𝑦=π‘₯βˆ’6π‘₯+𝐷 for some value of 𝐷.

We can find the value of 𝐷 by using the fact that the curve passes through the point (βˆ’2,βˆ’4). Substituting these values into the equation of our curve, we get βˆ’4=(βˆ’2)βˆ’6(βˆ’2)+π·βˆ’4=βˆ’8+12+𝐷𝐷=βˆ’8.

Finally, we substitute this value of 𝐷 into the equation of our curve to get our final answer: 𝑦=π‘₯βˆ’6π‘₯βˆ’8.

Let’s finish by recapping some of the points covered.

Key Points

  • Indefinite integration can be used to find the most general antiderivative of a function.
  • If we are told that dd𝑦π‘₯=𝑓(π‘₯), then we know that 𝑦=𝑓(π‘₯)π‘₯=𝑔(π‘₯)+.dC
  • 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.

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