Question Video: Finding the Integration of a Polynomial Involving Multiplying Two Brackets and Applying the Power Rule

Determine ∫ (βˆ’7π‘₯Β² βˆ’ 5)(βˆ’π‘₯Β³ + 1) dπ‘₯.

03:46

Video Transcript

Determine the integral of negative seven π‘₯ squared minus five multiplied by negative π‘₯ cubed plus one with respect to π‘₯.

In this question, we’re asked to evaluate the indefinite integral of the product of two polynomials. And we don’t know how to do this directly. However, we do know how to multiply both of these together by using the FOIL method. This would then turn our integrand into a polynomial which we know how to integrate term by term by using the power rule for integration. So we’ll start by multiplying these two polynomials together by using the FOIL method. First, we need to multiply the first two terms together. So by multiplying the first two terms of our polynomials together, we get negative seven π‘₯ squared multiplied by negative π‘₯ cubed.

And we can evaluate this and simplify by using our laws of exponents. It’s equal to seven π‘₯ to the fifth power. We’ll continue expanding this by using the FOIL method. Next, we need to multiply the outer two terms together. Doing this means we need to add on the term negative seven π‘₯ squared multiplied by one. Of course, multiplying by one doesn’t change our value, so we just get negative seven π‘₯ squared. Next, the FOIL method tells us to multiply the inner two terms together. This gives us negative five multiplied by negative π‘₯ cubed, which we can evaluate is equal to five π‘₯ cubed.

Finally, the FOIL method tells us we need to multiply the last two terms of our binomials together. So we have to add on the term negative five multiplied by one, which is of course equal to negative five. So we’ve rewritten our integral as the integral of seven π‘₯ to the fifth power minus seven π‘₯ squared plus five π‘₯ cubed minus five with respect to π‘₯. So now we’re just evaluating the integral of a polynomial.

And we know we can do this term by term by using the power rule for integration, which tells us 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 our constant of integration 𝐢. We add one to our exponent of π‘₯ and then divide by this new exponent. We want to apply this term by term.

Let’s start with the first term with seven π‘₯ to the fifth power. We see our exponent of π‘₯ is five. We want to add one to this exponent of five to give us a new exponent of six and then divide by this new exponent. This gives us seven π‘₯ to the sixth power divided by six. We want to do the same with our next term. We can see now our exponent of π‘₯ is two. We add one to our exponent of two to give us a new exponent of three and then divide by this new exponent. And remember, since we were subtracting seven π‘₯ squared, we need to subtract this value. This means we’re subtracting seven π‘₯ cubed over three.

We can do exactly the same with our third term; our exponent of π‘₯ is three. So once again, we add one to this exponent of three to give us a new exponent of four and then divide by this new exponent. This gives us five π‘₯ to the fourth power over four. And there are a few ways of integrating our last term. For example, we could rewrite negative five as negative five times π‘₯ to the zeroth power because remember π‘₯ to the zeroth power is equal to one. We could then evaluate the integral of this term by using the power rule for integration with our value of 𝑛 equal to zero.

Doing this would give us negative five π‘₯. We can also remember that the derivative of negative five π‘₯ with respect to π‘₯ is negative five. So negative five π‘₯ is an antiderivative of negative five. Finally, remember, we need to add a constant of integration. We’ll call this 𝐢. This gives us seven π‘₯ to the sixth power over six minus seven π‘₯ cubed over three plus five π‘₯ to the fourth power over four minus five π‘₯ plus 𝐢.

And we could leave our answer like this. However, since this is a polynomial, we’ll write this so our exponents of π‘₯ are decreasing. And so by swapping our second and third term around, we get our final answer. Therefore, we were able to show the integral of negative seven π‘₯ squared minus five times negative π‘₯ cubed plus one with respect to π‘₯ is equal to seven π‘₯ to the sixth power over six plus five π‘₯ to the fourth power over four minus seven π‘₯ cubed over three minus five π‘₯ plus 𝐢.

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