Video: Finding the Integration of an Exponential Function Multiplied by a Polynomial Using Integration by Parts

Determine ∫(3π‘₯ + 4)²𝑒^π‘₯dπ‘₯.

06:32

Video Transcript

Determine the integral of three π‘₯ plus four all squared multiplied by 𝑒 to the power of π‘₯ with respect to π‘₯.

We need to determine the value of an indefinite integral. We can see that our integrand is the product of two functions. We’ve seen a few different ways of dealing with integrals of this form. For example, we might be tempted to try integration by substitution. However, there’s no obvious substitution that would make this easier. Instead, we could try using integration by parts.

First, we recall integration by parts tells us the integral of 𝑒𝑣 prime with respect to π‘₯ is equal to 𝑒 times 𝑣 minus the integral of 𝑣 times 𝑒 prime with respect to π‘₯. So, integration by parts gives us a method of integrating the product of 𝑒 and 𝑣 prime with respect to π‘₯.

But, remember, we need to choose our function 𝑒. There’s a few different ways of choosing our function 𝑒. Usually, we pick the function which differentiates to give a simpler function. Using this logic, we see the derivative of 𝑒 to the power of π‘₯ is just equal to 𝑒 to the power of π‘₯. This doesn’t get any simpler. However, we know three π‘₯ plus four all squared is a quadratic. So, when we differentiate this with respect to π‘₯, we’ll get a linear function. And a linear function is more simple than a quadratic. So, this makes a good candidate for 𝑒.

Alternatively, we could’ve used the LIATE method. Since there are no logarithmic or inverse trigonometric functions in our integrand, we would choose the algebraic function three π‘₯ plus four all squared to be 𝑒. Both of these methods pick the same function. So, we’ll set 𝑒 to be three π‘₯ plus four all squared and 𝑣 prime to be 𝑒 to the power of π‘₯.

To use integration by parts, we also need expressions for 𝑒 prime and 𝑣 prime. Let’s start with finding 𝑒 prime. To find 𝑒 prime, we need to differentiate three π‘₯ plus four all squared. We could do this by using the chain rule or the general power rule. However, because we’re only squaring this linear function, we could do this by using the FOIL method or binomial expansion. We would get that 𝑒 is equal to nine π‘₯ squared plus 24π‘₯ plus 16. And now, we can differentiate this term by term by using the power rule for differentiation.

We need to multiply it by our exponents of π‘₯ and then reduce this exponent by one. This gives us that 𝑒 prime is equal to 18π‘₯ plus 24. Now, we need to find our function 𝑣. 𝑣 will just be an antiderivative of 𝑒 to the power of π‘₯. Well, we know the derivative of 𝑒 to the power of π‘₯ with respect to π‘₯ is just equal to 𝑒 to the power of π‘₯. So, we can just set 𝑣 to be 𝑒 to the power of π‘₯.

Now that we found expressions for 𝑒 prime and 𝑣, we can use integration by parts to attempt to evaluate our integral. Substituting in our expressions for 𝑒, 𝑣, 𝑒 prime, and 𝑣 prime, we get that our integral is equal to nine π‘₯ squared plus 24π‘₯ plus 16 all multiplied by 𝑒 to the power of π‘₯ minus the integral of 𝑒 to the π‘₯ times 18π‘₯ plus 24 with respect to π‘₯.

But now we can see a problem. We have an integral in this expression which we still can’t integrate directly. But now we can see something interesting. Our original integrand was a quadratic multiplied by 𝑒 to the power of π‘₯. However, we now have just a linear function multiplied by 𝑒 to the power of π‘₯. If we did this process one more time, then our linear function would just become a constant. Then, we could just evaluate our integral.

So, we’re gonna use integration by parts on the integral of 𝑒 to the power of π‘₯ times 18π‘₯ plus 24 with respect to π‘₯. This time, our function 𝑒 will be 18π‘₯ plus 24, and 𝑣 prime will be equal to 𝑒 to the power of π‘₯. And just as we did before, we can find expressions for 𝑒 prime and 𝑣 prime. 𝑒 prime will be the derivative of 18π‘₯ plus 24 with respect to π‘₯, which we know is 18. And 𝑣 will be 𝑒 to the power of π‘₯.

Now, all we need to do is substitute our expressions for 𝑒, 𝑣, 𝑒 prime, and 𝑣 prime into our formula for integration by parts. Doing this, we get 18π‘₯ plus 24 all multiplied by 𝑒 to the power of π‘₯ minus the integral of 𝑒 to the power of π‘₯ times 18 with respect to π‘₯. And we can rearrange our integrands to give us 18𝑒 to the power of π‘₯. And now, we can just evaluate this integral.

Negative one times the integral of 18𝑒 to the power of π‘₯ with respect to π‘₯ is equal to negative 18 times 𝑒 to the power of π‘₯ plus a constant of integration 𝐢. So, this gave us 18π‘₯ plus 24 all multiplied by 𝑒 to the power of π‘₯ minus 18𝑒 to the power of π‘₯ plus 𝐢. And before we substitute this expression back into our original integral, we can take out a factor of 𝑒 to the power of π‘₯. Taking out this common factor of 𝑒 to the power of π‘₯, we get 18π‘₯ plus 24 minus 18 all multiplied by 𝑒 to the power of π‘₯ plus 𝐢.

And of course, we can now simplify 24 minus 18 to give us six. So, we simplified this expression to give us 18π‘₯ plus six all multiplied by 𝑒 to the power of π‘₯ plus 𝐢. We can now substitute this expression back into our original integral. Doing this, we get nine π‘₯ squared plus 24π‘₯ plus 16 all multiplied by 𝑒 to the power of π‘₯ minus 18π‘₯ plus six times 𝑒 to the power of π‘₯ plus our constant of integration 𝐢.

We now want to distribute this negative one over our parentheses. But we see this will give us the term negative 𝐢. We could call this negative 𝐢. However, 𝐢 is our constant of integration. We can call this whatever we want. So, we could’ve originally called this negative 𝐢, so we can just write negative 𝐢 as 𝐢. This gives us nine π‘₯ squared plus 24π‘₯ plus 16 all multiplied by 𝑒 to the power of π‘₯ minus 18π‘₯ plus six times 𝑒 to the power of π‘₯ plus our constant of integration 𝐢.

And we could leave our answer like this. However, we see that our first two terms both share a factor of 𝑒 to the power of π‘₯. So, by taking out this common factor of 𝑒 to the power of π‘₯, we get nine π‘₯ squared plus 24π‘₯ plus 16 minus 18π‘₯ minus six all multiplied by 𝑒 to the power of π‘₯ plus 𝐢. And now, we can simplify this expression. We have 24π‘₯ minus 18π‘₯ is equal to six π‘₯. And 16 minus six is just equal to 10.

And then, by rewriting 𝑒 to the power of π‘₯ at the start of our first factor, we got 𝑒 to the power of π‘₯ times nine π‘₯ squared plus six π‘₯ plus 10 plus the constant of integration 𝐢. And this is our final answer. Therefore, by using integration by parts twice, we were able to determine that the integral of three π‘₯ plus four all squared times 𝑒 to the power of π‘₯ with respect to π‘₯ is equal to 𝑒 to the power of π‘₯ times nine π‘₯ squared plus six π‘₯ plus 10 plus the constant of integration 𝐢.

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