Question Video: Finding the Integration of a Function Involving Expanding a Square Bracket and Using the Power Rule | Nagwa Question Video: Finding the Integration of a Function Involving Expanding a Square Bracket and Using the Power Rule | Nagwa

Question Video: Finding the Integration of a Function Involving Expanding a Square Bracket and Using the Power Rule Mathematics • Second Year of Secondary School

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Determine ∫(4𝑥² + 6)² d𝑥.

04:52

Video Transcript

Determine the integral of four 𝑥 squared plus six all squared with respect to 𝑥.

The question is asking us to evaluate the integral of the square of a polynomial. And we don’t know how to do this directly. However, we do know how to distribute our square over our parentheses. This will then give us the integral of a polynomial which we can then evaluate.

There’s a few different ways of distributing our square over our parentheses. For example, we could use the binomial formula. However, we’re going to do this by using the FOIL method. First, the FOIL method tells us we need to multiply the first two terms together. This gives us four 𝑥 squared multiplied by four 𝑥 squared, which we can simplify to give us 16𝑥 to the fourth power.

Second, the FOIL method tells us we’ll want to multiply the outer two terms together. Doing this, we get four 𝑥 squared multiplied by six, which we can simplify to give us 24𝑥 squared. Third, the FOIL method tells us we want to multiply the inner two terms together. Which then gives us six times four 𝑥 squared, which we can calculate is equal to 24𝑥 squared. Finally, the FOIL method tells us we need to multiply the last two terms together. And this gives us six times six, which is equal to 36.

So, we’ve now rewritten our integrand as a polynomial. This means we can evaluate this integral. But, first, we can simplify our integrand. We can add the two terms of 24𝑥 squared together. Doing this, we get 48𝑥 squared. So, we now need to evaluate the integral of 16𝑥 to the fourth power plus 48𝑥 squared plus 36 with respect to 𝑥. This is just the integral of a polynomial. So, we’ll do this by splitting the integral of a sum into the sum of the integrals.

Doing this, we get the integral of 16𝑥 to the fourth power with respect to 𝑥 plus the integral of 48𝑥 squared with respect to 𝑥 plus the integral of 36 with respect to 𝑥. We’re now ready to evaluate each of these integrals by using the power rule for integration.

We recall the power rule for integration tells us if 𝑛 is not equal to negative one. Then for any constant 𝑎, the integral of 𝑎 times 𝑥 to the 𝑛th power with respect to 𝑥 is equal to 𝑎 times 𝑥 to the power of 𝑛 plus one divided by 𝑛 plus one plus the constant of integration 𝐶. We add one to our exponent of 𝑥 and then divide by this new exponent of 𝑥.

So, let’s evaluate each of these integrals separately. First, in the integral of 16𝑥 to the fourth power with respect to 𝑥, we can see our exponent of 𝑥 is four. To integrate this by using the power rule for integration, we want to add one to this exponent of four and then divide by this new exponent. This gives us 16𝑥 to the power four plus one divided by four plus one. And we then want to add our constant of integration. We’ll call this 𝐶 one. And, of course, we can simplify this since four plus one is equal to five.

We can do something similar to evaluate our second integral. We see the exponent of 𝑥 is two. And again, we add one to our exponent of two and then divide by this new exponent. This gives us 48 times 𝑥 to the power of two plus one divided by two plus one. And, of course, we need to add a constant of integration. This time, we’ll call this 𝐶 two. And just as we did before, we can simplify this. Two plus one is equal to three. In fact, we can then simplify this even further since 48 divided by three is equal to 16.

So, now, we need to evaluate our third and final integral, the integral of 36 with respect to 𝑥. However, there’s a problem. We don’t seem to be able to use the power rule for integration in this case since there’s no exponent of 𝑥. However, remember that 𝑥 to the zeroth power is equal to one. So, we can rewrite our integral of 36 as 36 times 𝑥 to the zeroth power. So, now, we need to evaluate the integral of 36𝑥 to the zeroth power with respect to 𝑥. We can do this by using the power rule for integration. Our exponent of 𝑥 is equal to zero.

And now, we evaluate this integral using the same method. We add one to our exponent of zero and then divide by this new exponent. This gives us 36𝑥 to the power of zero plus one divided by zero plus one. And, of course, we need to add a constant of integration. We’ll call this 𝐶 three. And just as we did before, we can simplify this since zero plus one is equal to one. In fact, we can then simplify this further since 36 divided by one is just equal to 36 and 𝑥 to the first power is just equal to 𝑥. This gives us 36𝑥.

So, we’ve now evaluated all three of our integrals. There’s one more thing we can do to simplify this expression. Since 𝐶 one, 𝐶 two, and 𝐶 three are all constants of integration, we can combine all three of these into one new constant of integration which we will call 𝐶. And this gives us our final answer.

Therefore, we were able to show the integral of four 𝑥 squared plus six all squared with respect to 𝑥 is equal to 16𝑥 to the fifth power divided by five plus 16𝑥 cubed plus 36𝑥 plus the constant of integration 𝐶.

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