Question Video: Evaluating the Definite Integration of a Power Function with Fraction Exponent | Nagwa Question Video: Evaluating the Definite Integration of a Power Function with Fraction Exponent | Nagwa

# Question Video: Evaluating the Definite Integration of a Power Function with Fraction Exponent Mathematics • Third Year of Secondary School

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Evaluate ∫_(1)^(27) 4𝑥⁻²ᐟ³ 𝑑𝑥.

04:44

### Video Transcript

Evaluate the integral of four 𝑥 to the power of negative two over three and that’s with the limits 27 and one.

So what we actually have in the question is a definite integral. And we have a definite integral because actually we know the limits of the integration that we want to carry out. And we can use that because that’s actually gonna help us to find an area under a graph.

So what I’ve done is I’ve drawn a little sketch to help us understand that. And what it means is that if we actually have a function 𝑦 equals 𝑓 𝑥, then if we want to find the area underneath that function on a graph between the two limits 𝑎 and 𝑏, then all we do is we actually find the integral of the function between 𝑏 and 𝑎.

However, how do we actually calculate the value of what this is going to be? Well, to do this, what we actually use is something called the definite integral formula. And what this tells us that if we have the integral of 𝑓 𝑥 — I’m gonna call that 𝐹 𝑥 with a capital 𝐹 — then if we’re looking to find the definite integral between two limits, 𝑏 and 𝑎, of our integrant, which is what we call it the 𝑓 𝑥 function that we’re actually integrating, then this is gonna be equal to our integral with the value 𝑏, so our upper limit minus the integral with the value 𝑎, which is our lower limit. And they’re both substituted in for 𝑥.

Okay, so we’re now gonna use this to actually evaluate our expression. And what it means in practice is that we’re actually going to integrate our expression. Then we’re actually going to substitute in the values for the upper and lower limits into the integral. And then we’re going to subtract the value for the lower limit from the value from the upper limit.

Okay, let’s go ahead and do that. So step one, we’re actually going to integrate four 𝑥 to the power of negative two over three, which is going to give us four 𝑥 to the power of a third over a third. And then that’s all again with our limits 27 and one. Just a quick reminder of how we actually integrated that term, well, what we did is we got four 𝑥 and then to the power of negative two over three. And then we’ve added one to the exponent. And that’s like adding three over three. So it gave us one over three. And then we divided by the new exponent, so one over three.

Okay, great! Then let’s move on to the next step and actually simplify this. Well, if we simplify, we’re just gonna get 12𝑥 to the power of a third, and then again with the same limits 27 and one. And just to kind of remind us how we did that, well, we had four 𝑥 to the power of a third. And that was divided by a third, so divided by a fraction.

Well, we know that, actually, if we’re gonna divide by a fraction, it’s the same as multiplying by the reciprocal of that fraction. So it was the same as four 𝑥 to the power of a third multiplied by three, which gives us 12𝑥 to the power of a third. Okay, so now to actually find out what the value of our definite integral is, we need to actually substitute in our upper and lower limits into our integral.

So first of all, in brackets, we actually have 12 multiplied by 27 to the power of a third. So I’ve actually substituted in our upper limit of 27 in for our value of 𝑥. And then we’re actually gonna minus 12 multiplied by one to the power of a third. And the reason we do that is because, actually, we’ve substituted in our lower value or our lower limit, which is one. So we get 12 multiplied by one to the power of a third.

So now to actually simplify this, what we’re gonna use is this exponent rule that tells us that 𝑥 to the power of one over 𝑎 is the same as the 𝑎th root of 𝑥. So this is gonna be equal to 12 multiplied by the cube root of 27 minus 12 multiplied by the cube root of one, which is 12 multiplied by three minus 12.

We got that because three is the cube root of 27, which is equal to 36 minus 12. So therefore, we can actually say that the definite integral of four 𝑥 to the power of negative two over three with the limits 27 and one has the value of 24. And to remind us of what we mentioned at the beginning, what this means in practice is that if we were to draw the graph of four 𝑥 to the power of negative two over three, then the area under the graph between the limits of 27 and one will actually be equal to 24.

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