Question Video: Determining the Definite Integration of a Function Involving an Exponential Function

Determine ∫_(0)^(1) (2𝑒^(9π‘₯) βˆ’ π‘₯) dπ‘₯.

03:15

Video Transcript

Determine the integral of two 𝑒 to the nine π‘₯ minus π‘₯ with respect to π‘₯ between the limits of one and zero.

We’re actually going to be integrating two different functions here. We’re going to integrate two 𝑒 to the power of nine π‘₯ with respect to π‘₯ and negative π‘₯ with respect to π‘₯. Since negative π‘₯ is the same as negative one π‘₯, we can split this integral up. It’s the same as the integral between the limits of one and zero of two 𝑒 to the nine π‘₯ with respect to π‘₯ minus one times the integral between the limits of one and zero of π‘₯ with respect to π‘₯.

And in fact, we can even simplify this first integral a little bit by taking out a factor of two. We’re allowed to do this. It just means that at the end of our calculation, we’re going to multiply everything by two rather than doing it at the beginning. So, how do we integrate 𝑒 to the power of nine π‘₯?

Well, we can think of integration as the opposite of differentiation. And we know that when we differentiate 𝑒 to the power of π‘₯ with respect to π‘₯, we simply get 𝑒 to the power of π‘₯. So, when we integrate 𝑒 to the power of π‘₯ with respect to π‘₯, we also get 𝑒 of the power of π‘₯ plus that constant of integration 𝑐. Well, we can extend this and say that if we differentiate 𝑒 to the power of some constant π‘˜ multiplied by π‘₯, we get π‘˜ times 𝑒 to the π‘˜π‘₯. And this means the integral of 𝑒 to the π‘˜π‘₯ with respect to π‘₯ is one over π‘˜ 𝑒 to the π‘˜π‘₯ plus that constant of integration 𝑐.

That means the integral of 𝑒 to the power of nine π‘₯ with respect to π‘₯ is a ninth 𝑒 to the power of nine π‘₯. So, two times the integral of 𝑒 to the power of nine π‘₯ is two times a ninth 𝑒 to the power of nine π‘₯. But we’re actually gonna evaluate a ninth 𝑒 to the power of nine π‘₯ between the limits of one and zero before multiplying by two. The integral of π‘₯ is a little bit more straightforward. It’s π‘₯ squared divided by two.

Our next step is to evaluate each of these expressions between one and zero. Evaluating a ninth 𝑒 to the power of nine π‘₯ between the limits one and zero, and we get a ninth 𝑒 to the nine times one minus a ninth 𝑒 to the nine times zero. And if we evaluate π‘₯ squared divided by two between one and zero, we get one squared divided by two minus zero squared divided by two. Zero squared divided by two is simply zero. And nine multiplied by zero is also zero. So, we have a ninth 𝑒 to the power of zero. And since 𝑒 to the power of zero is one, this just becomes one-ninth.

And simplifying, we see that we have two times a ninth 𝑒 to the power of nine minus a ninth all minus one-half. We’ll distribute this first set of parentheses by multiplying each term one-ninth 𝑒 to the nine and negative a ninth by two. And our final step is going to be to evaluate negative two-ninths minus a half.

To evaluate that, we’ll need to make the denominators the same. The lowest common denominator of nine and two is 18. So, we need to multiply both the numerator and the denominator of two-ninths by two, and the numerator and the denominator of one-half by nine. Two-ninths is, therefore, equivalent to four eighteenths, and a half is equivalent to nine-eighteenths. Negative four eighteenths minus nine eighteenths is negative thirteen eighteenths. So, the integral of two 𝑒 to the nine π‘₯ minus π‘₯ with respect to π‘₯ between the limits of one and zero is two-ninths 𝑒 to the power of nine minus thirteen eighteenths.

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