Question Video: Evaluating the Definite Integration of a Polynomial | Nagwa Question Video: Evaluating the Definite Integration of a Polynomial | Nagwa

# Question Video: Evaluating the Definite Integration of a Polynomial Mathematics • Third Year of Secondary School

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Evaluate β«_(0) ^(1) (8π£β· + 12π£Β³ + 3)ππ£.

03:21

### Video Transcript

Evaluate the integral one, zero of eight π£ to power of seven plus 12π£ cubed plus three ππ£.

Okay, so if we want to evaluate this definite integral, the first stage is to actually integrate our expression. And to remind us how to do that, we have the general form which says that the integral of ππ’ to the power of π ππ’ is equal to ππ’ to the power of π plus one over π plus one plus πΆ.

So what does this mean in practice? Well, in practice what it means is we actually increase the exponent of our term by one and then divide by the new exponent. Itβs also worth noting as weβre dealing with a definite integral in our question, we donβt need to include the plus πΆ at the end.

Okay, great! So now letβs get on and integrate our expression. So our first term is going to be eight over eight π£ to the power of eight. And this is because we raise the exponent by one, so I went from seven to eight, and then divide it by this new exponent, so divided by eight. And then we have 12 over four π£ to the power of four. And then our final term is three π£.

Okay, great! So weβve integrated. What do we do now? Well we can actually take a look at this, because if we have the integral of ππ’ ππ’ equals ππ’ plus πΆ, then we can actually say that the integral of π, π ππ’ ππ’ is equal to ππ’, and then with the limits π, π which is the stage that weβre at because actually weβve integrated and weβve got our limits, where π is our upper limit and π is our lower limit.

Then this is going to be equal to ππ minus ππ. And what this means is actually the value of π at the upper limit minus the value of π at the lower limit. Okay, great! Now we know what to do. Letβs do this next step with our expression. In order to complete this next step, what weβre going to do is first of all substitute one in for π£.

But just before we do, weβre actually going to simplify our expression one stage further just to make life easier, which gives us π£ to power of eight plus three π£ to the power of four plus three π£ and then our limits one, zero. So now we can actually do the final stage and we can substitute in one for π£ and zero for π£ to get our upper and lower limit values.

So weβre gonna get one to power of eight plus three multiplied by one to the power four plus three multiplied by one, and then minus zero to power of eight plus three times zero to the power four plus three times zero. We can actually see that our lower limit is just going to be zero because weβve been multiplying by zero throughout.

So weβre gonna get one plus three plus three, which is equal to seven. So therefore, we can say that if you evaluate the integral of eight π£ to the power of seven plus 12π£ cubed plus three ππ£ with the limits one, zero, then the answer is equal to seven.

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