Video: Evaluating the Definite Integration of a Polynomial

Evaluate ∫_(0)^(1) ((4/5)𝑡³ + (3/4)𝑡² − (2/3)𝑡) 𝑑𝑡.

04:31

Video Transcript

Evaluate the definite integral of four over five 𝑡 cubed plus three over four 𝑡 squared minus two over three 𝑡 with the limits one and zero.

When dealing with a definite integral, we actually have a general form that can help us find out what the value of it is. So we know that if we’re finding the definite integral of a function with limits 𝑏 and 𝑎, then this is gonna be equal to the integral with the limits 𝑏 and 𝑎. And what that actually means is the integral with the upper limit substituted in — so 𝑏 substituted in — and then minus the integral with the lower limit, which is 𝑎 substituted in. And these are both substituted in for our 𝑥-values.

So now, we know actually what we need to do. Let’s evaluate our expression and actually find out the value of our definite integral. So our first step is to actually integrate four over five 𝑡 cubed plus three over four 𝑡 squared minus two over three 𝑡. Well, our first term is going to be four 𝑡 to the power of four over five multiplied by four. And we got that because what we’ve done is we’ve actually raised the exponent of our 𝑡 term by one — so from three to four. So it gave us four 𝑡 to the power of four.

And then, what we actually do is we divide by the new exponent. But because we already had four-fifths of 𝑡 cubed, then it means we’re actually gonna multiply five by the new exponent, which is four, on the denominator. So it’s gonna give us four 𝑡 to the power of four over five multiplied by four. And I’ve left the denominator in this form just because it’s gonna be easier in a second when I show you how we’d actually simplify this particular example.

And then, our second term is gonna be three 𝑡 cubed over four multiplied by three. Again using the same method, we raise the exponent by one — so from two to three. And then, we divide it by the new exponent. So that gave us four multiplied by three on the denominator because we already had four in the denominator because it was three-quarters 𝑡 squared. And our final term is two 𝑡 squared over three multiplied by two. And again, we did this in the exact same way. We raise the exponent from one to two and then we divide it by the new exponent.

Okay, great, so now what we’re gonna do is actually simplify this. Well, our first term is 𝑡 to the power of four over five. And as you can see, this is because we actually what we did was divide it through by four. And we’ve divided the numerator and denominator by four. They actually cancel each other out. So we’ve got 𝑡 to the power of four over five. Then, this is plus 𝑡 cubed over four. Again, same reasons, if we divide top and bottom or numerator and denominator by three, we’re just left with 𝑡 cubed over four. And then, our final term is 𝑡 squared over three. Again, using the same method, we’ve cancelled the twos out by dividing the numerator and denominator by two. So it leaves us with 𝑡 squared over three.

Okay, great, so we’re now at the stage where we can actually look to substitute in our values for our limits. And if we take a look over at the general form, we can see that what we want to do is actually substitute the value for the highest limit first. And then, what we’re gonna do is subtract the value when the lowest limit is substituted in from this. So then, we get one to the power of four over five plus one to the power of three over four minus one to the power of two over three minus- and then I wouldn’t usually write this, but I’ve just put it down just so I can show the complete working. We’ve got nought to the power of four over five plus nought to the power of three over four minus nought to the power of two over three, which is just going to be zero. So therefore, it’s gonna be equal to a fifth plus a quarter minus a third.

And then, what we do just in case it’s a noncalculator question, we actually make sure they’ve got the same denominator. So a fifth would actually become 12 over 60. And we’ve chosen 60 because actually it’s the lowest common multiple of five, four, and three. And we got 12 over 60 because 12 multiplied by five gives us 60. Then, our next one will be 15 over 60 because a quarter is 15 over 60. And then, finally, the last fraction is 20 over 60.

Okay, great, so now we’ve got the same denominator. We can find the final answer. So therefore, we can say that the definite integral of four over five 𝑡 cubed plus three over four 𝑡 squared minus two over three 𝑡 with the limits one and zero is equal to seven over 60.

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