Lesson Video: Applications of Indefinite Integration Mathematics

In this video, we will learn how to use indefinite integration to express a function given its rate of change.

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Video Transcript

Applications of Indefinite Integration

In this video, we’re going to learn how we can use indefinite integration to express the function given an expression for its rate of change. And we’ll see several different ways we can apply this in different situations. To do this, we’re going to first need to recall what we mean by the indefinite integral of a function.

For example, let’s consider the equation 𝑦 is equal to π‘₯ squared. We’re going to integrate π‘₯ squared with respect to π‘₯. And we know how to do this. We add one to our exponent of two giving us a new exponent of three, and then we divide by this new exponent. This gives us π‘₯ cubed over three, and then we add our constant of integration 𝐢. But we need to remember exactly why we’re calculating this expression. Remember, when we first defined integration, we wanted to do this to find antiderivatives of different expressions. So we defined integration to be the reverse process of differentiation. In other words, what we’re calculating when we integrate π‘₯ squared is an antiderivative of π‘₯ squared. And in fact, we call this the most general antiderivative because it’s an antiderivative for any value of our constant 𝐢. And to see exactly what this means, let’s differentiate this expression with respect to π‘₯.

If we wanted to differentiate π‘₯ cubed over three plus our constant of integration 𝐢 with respect to π‘₯, we would do this term by term. To differentiate our first term, we want to multiply it by the exponent of π‘₯ and then reduce this exponent by one. Of course, we can see this is going to give us π‘₯ squared. Then, 𝐢 is a constant, so its rate of change with respect to π‘₯ is going to be zero. It doesn’t vary as our value of π‘₯ varies. So the derivative of this expression is exactly what we started with, π‘₯ squared. And of course, this is true for any value of our constant 𝐢. And this leads us to a very useful property. What if instead of having 𝑦 is equal to π‘₯ squared, we have instead started with d𝑦 by dπ‘₯ is equal to π‘₯ squared?

So now we start with d𝑦 by dπ‘₯ is equal to π‘₯ squared. The derivative of 𝑦 with respect to π‘₯ is equal to π‘₯ squared. We do exactly the same lines of working out and we end up with the derivative of this expression is equal to π‘₯ squared. So we started with the derivative of 𝑦 with respect to π‘₯ is equal to π‘₯ squared and we ended up with the derivative of this expression with respect to π‘₯ is equal to π‘₯ squared. In other words, we found an expression for 𝑦 up to our constant of integration 𝐢. And because we found this expression by integrating, you’ll often see this as 𝑦 is equal to the integral of our slope with respect to π‘₯.

So now if we’re given the slope of a function, we know we can integrate this with respect to π‘₯ to try and find our original equation. And as long as we know how to integrate this, we’ll be able to find an expression for 𝑦 up to our constant of integration 𝐢. The only question that remains is how we’re going to find the value of 𝐢. To find the value of 𝐢, we’re going to need to know a little bit more information. Usually, when we’re given the slope of a curve, we’ll be given a point on the curve. For example, we might be told that the curve 𝑦 is equal to 𝑓 of π‘₯ passes through the point one, one. This is often called the initial value. Then, if we’ve shown that 𝑦 is equal to π‘₯ cubed over three plus our constant of integration 𝐢, we can substitute this point into our curve.

Since our curve passes through the point one, one, we know when π‘₯ is equal to one, 𝑦 must be equal to one. Substituting these values into our curve, we get one is equal to one cubed over three plus our constant of integration 𝐢. Simplifying and subtracting one-third from both sides of the equation, we get that 𝐢 must be equal to two over three. Then we can just write this value of 𝐢 into the equation for our curve. And this gives us our equation for our curve, 𝑦 is equal to π‘₯ cubed over three plus two over three. Therefore, what we’ve actually shown is if the slope of a curve is equal to π‘₯ squared and the curve 𝑦 is equal to 𝑓 of π‘₯ passes through the point one, one, then our curve must have the equation 𝑦 is equal to π‘₯ cubed over three plus two-thirds. And we’ll be able to do this in general, provided we can integrate the function in the slope. Let’s now see an example of how we might do this in practice.

Find the equation of the curve that passes through the point negative two, one given that the gradient of the tangent to the curve is negative 11π‘₯ squared.

In this question, we want to find the equation of a curve. And to do this, we’re given some information about our curve. We’re told that our curve passes through the point negative two, one and the gradient of the tangent to this curve is equal to the function negative 11π‘₯ squared. So let’s break down the two pieces of information we’re given about our curve. First, if we wanted to write the equation of this curve in the form 𝑦 is equal to 𝑓 of π‘₯, then saying that our curve passes through the point negative two, one means when π‘₯ is equal to negative two, 𝑦 should be equal to one.

So we can substitute these values into the equation for our curve. We should have that one is equal to 𝑓 evaluated at negative two. Next, we’re told the gradient of the tangent to our curve is negative 11π‘₯ squared. And remember, we know the gradient of the tangent to our curve at a value of π‘₯ is as rate of change of 𝑦 with respect to π‘₯. In other words, this is telling us d𝑦 by dπ‘₯ is equal to negative 11π‘₯ squared. So we need to find the equation of a curve given its slope and a point on the curve. And to do this, we’re going to need to find an antiderivative of negative 11π‘₯ squared. And we know how to find antiderivatives by using integration. The integral of negative 11π‘₯ squared with respect to π‘₯ will give us the most general antiderivative of negative 11π‘₯ squared.

And remember, the reason we’re finding this is we know the derivative of our expression for 𝑦 with respect to π‘₯ has to be equal to negative 11π‘₯ squared. So we need to integrate negative 11π‘₯ squared with respect to π‘₯. We can do this by using the power rule for integration. We recall this tells us for any real constants π‘Ž and 𝑛, where 𝑛 is not equal to negative one, the integral of π‘Ž times π‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘Ž times π‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant of integration 𝐢. We add one to our exponent and then divide by this new exponent.

In our case, the value of π‘Ž is negative 11 and the value of 𝑛 is two. So we add one onto our exponent of two to get an exponent of three and then divide by this new exponent. We get negative 11π‘₯ cubed over three plus a constant of integration 𝐢. And of course, the derivative of this expression with respect to π‘₯ is equal to negative 11π‘₯ squared. So far, for our curve 𝑦 is equal to negative 11π‘₯ cubed over three plus a constant of integration 𝐢, we have d𝑦 by dπ‘₯ is equal to negative 11π‘₯ squared. However, remember, we still need this curve to pass through the point negative two, one. And to do this, we need to substitute π‘₯ is equal to negative two and 𝑦 is equal to one in to our curve.

Substituting π‘₯ is equal to negative two and 𝑦 is equal to one into the equation for our curve, we get one is equal to negative 11 times negative two cubed all over three plus our constant of integration 𝐢. Now all we need to do is solve this equation for 𝐢. Rearranging this equation, we see that 𝐢 should be equal to negative 85 divided by three. Now all we need to do is use this value of 𝐢 in the equation for our curve, and this gives us our final answer. Therefore, we were able to show the equation of the curve that passes through the point negative two, one and which has the gradient of its tangent line at π‘₯ given by negative 11π‘₯ squared has the equation 𝑦 is equal to negative 11 over three times π‘₯ cubed minus 85 over three.

Let’s now see an example where the gradient is given by a more complicated function.

Find the equation of the curve given the gradient of the tangent is five times the sin squared of π‘₯ over two and the curve passes through the origin.

In this question, we’re asked to find the equation of a curve. And to do this, we’re given some information about our curve. First, we’re told that the gradient of the tangent line to our curve at π‘₯ is given by five times the sin squared of π‘₯ over two. Next, we’re also told that our curve passes through the origin. To answer this question, let’s start by saying we can write the equation of our curve in the form 𝑦 is equal to some function 𝑓 of π‘₯. Then we need two things to be true about this equation. First, we’re told that our curve must pass through the origin. And remember, the origin is the point with both π‘₯- and 𝑦-coordinate equal to zero.

And if our curve 𝑦 is equal to 𝑓 of π‘₯ passes through the origin, then this equation must be satisfied when π‘₯ is equal to zero and 𝑦 is equal to zero. In other words, we must have zero is equal to 𝑓 evaluated at zero. This is not the only piece of information we’re given. We’re also told the gradient of the tangent to our curve. Remember, though, the gradient of the tangent to our curve, our value of π‘₯, is the rate of change of 𝑦 with respect to π‘₯. In other words, we’re told that d𝑦 by dπ‘₯ must be equal to five times the sin squared of π‘₯ over two. And it’s also worth pointing out, we could’ve wrote this down as 𝑓 prime of π‘₯. It’s personal preference which one you would use.

So to find the equation of this curve, we need both of these pieces of information to be true. In particular, we’re first looking for a function which differentiates to give us five times the sin squared of π‘₯ over two. In other words, we’re looking for an antiderivative of this function. And we know how to find antiderivatives by using integration. If we integrate this with respect to π‘₯, we’ll find an antiderivative of this. In other words, 𝑦 will be equal to the integral of five times the sin squared of π‘₯ over two with respect to π‘₯ up to a constant of integration. However, we can see a problem. It’s very difficult to integrate the sin squared function. So we’re going to want to simplify this expression first. And to do this, we’re going to use the double-angle formula for cos.

Recall one version of this tells us the cos of two πœƒ is equivalent to one minus two times the sin squared of πœƒ. And this will be true for any value of πœƒ. We want to use this to find an expression for the sin squared of π‘₯ over two. And to do this, we’re going to need to substitute πœƒ is equal to π‘₯ over two. Doing this, we get the cos of two times π‘₯ over two is equivalent to one minus two sin squared of π‘₯ over two. And of course, we can simplify this, since two divided by two is equal to one. Next, we’ll rearrange this equation; we’ll add two sin squared of π‘₯ over two to both sides and subtract the cos of __ from both sides. This gives us two sin squared of π‘₯ over two is equivalent to one minus the cos of π‘₯.

There’s a few different options of what we could do next. Since we’re integrating five sin squared of π‘₯ over two, we should try and make our coefficient five. And to do this, we’ll multiply both sides of the equivalence through by five over two. And if we do this and simplify, we get five sin squared of π‘₯ over two is equivalent to five over two minus five over two cos of π‘₯. And we can see that this expression is much easier to integrate. So we’ll substitute this into our integral. This gives us that 𝑦 is equal to the integral of five over two minus five over two times the cos of π‘₯ with respect to π‘₯ up to our constant of integration.

Now we could just evaluate this integral term by term. First, five over two is a constant. When we integrate this with respect to π‘₯, we get five over two times π‘₯. Next, to integrate our second term, we recall the following rule. For any real constant π‘Ž, the integral of negative π‘Ž cos of π‘₯ with respect to π‘₯ is negative π‘Ž sin of π‘₯ plus the constant of integration 𝐢. We can use this with π‘Ž equal to negative five over two to get the integral of this term. So we have 𝑦 is five over two π‘₯ minus five over two times the sin of π‘₯ plus our constant of integration 𝐢. Now all that’s left to do is find the value of 𝐢. And to do this, we need to use the fact that zero is equal to 𝑓 evaluated at zero.

In other words, because our curve passes through the origin, we can just substitute π‘₯ is equal to zero and 𝑦 is equal to zero into our expression for the curve. And this equation must be satisfied. This gives us that zero is equal to five over two times zero minus five over two times the sin of zero plus 𝐢. Of course, five over two times zero is equal to zero and the sin of zero is also equal to zero. So this entire expression just simplifies to give us that 𝐢 is equal to zero. So we’ve shown that 𝐢 must be equal to zero. All we need to do now is substitute this value of 𝐢 into our expression for our curve. And by substituting 𝐢 is equal to zero, we get our final answer. The equation of this curve was 𝑦 is equal to five over two times π‘₯ minus five over two times the sin of π‘₯ over two.

Let’s now see an example involving exponential functions.

Given that the slope at the point π‘₯, 𝑦 is three 𝑒 to the power of three π‘₯ and 𝑓 evaluated at zero is negative three, determine 𝑓 evaluated at negative three.

In this question, we need to determine what 𝑓 evaluated at negative three is. And we’re told some information about our function 𝑓. We’re told that the slope at the point π‘₯, 𝑦 is given by three 𝑒 to the power of three π‘₯. And we’re also told that 𝑓 evaluated at zero is negative three. So we’re given two pieces of information about 𝑓 of π‘₯. First, we’re told the slope is three 𝑒 to the power of three π‘₯. And another way of saying this is 𝑓 prime of π‘₯ is equal to three 𝑒 to the power of three π‘₯. So to find 𝑓 of π‘₯, we need to find an antiderivative of 𝑓 prime of π‘₯. And we know how to do this by using integration.

We can find an antiderivative of 𝑓 prime of π‘₯ by integrating it with respect to π‘₯. We have 𝑓 of π‘₯ will be equal to the integral of three 𝑒 to the power of three π‘₯ with respect to π‘₯ up to a constant of integration. And to evaluate this integral, we need to recall the following. For any real constant π‘Ž, the integral of 𝑒 to the power of π‘Žπ‘₯ with respect to π‘₯ is equal to one over π‘Ž times 𝑒 to the power of π‘Žπ‘₯ plus a constant of integration 𝐢. We need to divide by the coefficient of π‘₯ in our exponent. And in our function, that’s equal to three. So we get three multiplied by one-third times 𝑒 to the power of three π‘₯. And remember, we need to add a constant of integration 𝐢.

Of course, we can simplify this. Three multiplied by one-third is just equal to one. So this simplifies to give us 𝑓 of π‘₯ is equal to 𝑒 to the power of three π‘₯ plus 𝐢. Now we want to find the value of 𝐢. And to do this, we need to use the fact that 𝑓 evaluated at zero is equal to negative three. So we substitue π‘₯ is equal to zero. We know 𝑓 of zero is negative three, and this is equal to 𝑒 to the power three times zero plus 𝐢. And now we just solved this equation for 𝐢. 𝑒 to the zeroth power is just equal to one. And then we rearrange to get 𝐢 is equal to negative four. So if 𝐢 is equal to negative four, we can substitute this into our expression for 𝑓 of π‘₯.

We have that 𝑓 of π‘₯ is equal to 𝑒 to the power of three π‘₯ minus four. The question wants us to find 𝑓 evaluated at negative three. So we substitute π‘₯ is equal to negative three into this expression. We get 𝑒 to the power of three times negative three minus four. And if we evaluate this expression and rearrange, we get our final answer of negative four plus one over 𝑒 to the ninth power.

Let’s now see an example of a question involving the maxima and minima of a curve.

Find the local maximum and minimum values of the curve that passes through the point negative one, seven where the gradient of the tangent is six times π‘₯ squared plus four π‘₯ plus three.

In this question, we need to find the local maximum and minimum values of a curve, but we’re not given the equation for our curve directly. Instead, we’re only given information about our curve. We’re told that our curve passes through the point negative one, seven. And we’re also given the gradient of the tangent line at the point π‘₯, 𝑦. It’s given by six times π‘₯ squared plus four π‘₯ plus three. Of course, to find the local maximum and minimum values of a function, we’re going to need to know its output at these values. So we’re going to want to find an expression for our function. And to do this, we’re going to need to use the two pieces of information we’re given about our curve.

First, we’re given an expression which represents the gradient of the tangent lines to our curve at a value of π‘₯. And we recall this is exactly the same as saying the rate of change of 𝑦 with respect to π‘₯ is equal to this expression. In other words, we have d𝑦 by dπ‘₯ is equal to six times π‘₯ squared plus four π‘₯ plus three. But this is not the only piece of information we’re given. We’re also told that our curve passes through the point negative one, seven. So when π‘₯ is equal to negative one, we must have that 𝑦 is equal to seven. We’re given an expression for the gradient of our curve.

So differentiating our curve with respect to π‘₯, we get the following expression. This means we want to find an antiderivative of this expression. And we know how to do this by using integration. We must have that 𝑦 is equal to the integral of six times π‘₯ squared plus four __ plus three with respect to π‘₯ up to a constant of integration because this will give us the most general antiderivative of this expression. There’s a few different ways of solving this. We’re going to distribute six over our parentheses. Doing this, we get the integral of six π‘₯ squared plus 24π‘₯ plus 18 with respect to π‘₯. And now this is just the integral of a quadratic. We can do this term by term by using the power rule for integration.

We recall for any real constants π‘Ž and 𝑛, where 𝑛 is not equal to negative one, the integral of π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘Ž times π‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant of integration 𝐢. We add one to our exponent of π‘₯ and then divide by this new exponent. Applying this to six π‘₯ squared, we add one to the exponent of two giving us a new exponent of three and then divide by the exponent of three. Doing this and simplifying, we get two π‘₯ cubed. We’ll do the same for 24π‘₯, which we can write as 24π‘₯ to the first power. We add one to our exponent of one and then divide by this new exponent. This gives us 12π‘₯ squared. Finally, we could do the same for 18 by writing it as 18 times π‘₯ to the zeroth power. However, we can just write the antiderivative of this as 18π‘₯. And don’t forget, we need to add our constant of integration 𝐢.

So now we found an expression for the curve given to us in the question up to a constant of integration 𝐢. We now want to find the value of our constant 𝐢. And to do this, we’re going to need to use the fact that our curve passes through the point negative one, seven. If our curve passes through this point, we can substitute these into the equation for our curve. And then the equation has to be true for our curve to pass through this point. Substituting π‘₯ is equal to negative one and 𝑦 is equal to seven into the equation for our curve, we get seven is equal to two times negative one cubed plus 12 times negative one squared plus 18 times negative one plus 𝐢.

Now by simplifying this expression and solving for 𝐢, we get that 𝐢 is equal to 15. And if 𝐢 is equal to 15, we can substitute this into the expression for our curve. So by clearing some space and using 𝐢 is equal to 15, we have the following expression for our curve. But remember, we need to find the local maximum and minimum values of this curve. And to find this, we need to recall the following piece of information. In polynomials, our local extrema will occur at the turning points, in other words, when the gradient is equal to zero. And we’re given an expression for the gradient.

So to find our local extrema, we need to solve six times π‘₯ squared plus four π‘₯ plus three is equal to zero. And to do this, all we need to do is factor our quadratic. We see this factors to give us π‘₯ plus one times π‘₯ plus three. Solving this is equal to zero, we get π‘₯ is negative one or π‘₯ is negative three. To find the values of these, we need to substitute these into the equation for our curve. Substituting negative one into our curve, we get 𝑦 is equal to seven. Similarly, when π‘₯ is equal to negative three, we see that 𝑦 is equal to 15. We now need to determine which of these is a maximum and which is a minimum. To do this, we notice our curve is a cubic with a positive leading coefficient. It will look something like this. In fact, we know it has two turning points, and we know the coordinates of these two turning points. So we’ve shown the local maximum value is 15 and the local minimum value is seven.

The key point of this video is if we’re given the slope of a function d𝑦 by dπ‘₯, we can find an equation for 𝑦 by using integration.

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