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

Evaluate ∫_(0)^(1) (π‘₯^(𝑒) + 2𝑒^(π‘₯)) dπ‘₯.

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Video Transcript

Evaluate the integral of the function π‘₯ to the power of 𝑒 plus two multiplied by 𝑒 to the π‘₯ with respect to π‘₯ between the limits of one and zero.

We can start by splitting our integral into two simple integrals by using the fact that the integral of the sum of two functions is equal to the sum of the integrals of those two functions. This gives us two integrals which we need to calculate, the integral of π‘₯ to the 𝑒 and the integral of two multiplied by 𝑒 to the π‘₯, both with respect to π‘₯.

We can notice that 𝑒 is a constant, and we know that for a constant 𝑛, we can integrate π‘₯ to the 𝑛th power by using the power rule for integration. Which says that the integral of π‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘₯ to the 𝑛 plus one over 𝑛 plus one plus our constant of integration.

In our case, we have that 𝑛 is equal to 𝑒. So, we will get π‘₯ to the 𝑒 plus one divided by 𝑒 plus one plus our constant of integration. However, since we are evaluating a definite integral, we can remove the constant of integration since it will cancel out when we evaluate the bounds of our integral. We can recall that the integral of a constant π‘˜ multiplied by 𝑓 of π‘₯ with respect to π‘₯ is equal to π‘˜ multiplied by the integral of 𝑓 of π‘₯ with respect to π‘₯. So, we can use this to take the coefficient of two out of our integral.

We know that if we differentiate 𝑒 to the π‘₯, then we’d just get 𝑒 to the π‘₯. If we were then to integrate both sides of this equation with respect to π‘₯, we would have that the integral of the derivative of 𝑒 to the π‘₯ is equal to the integral of 𝑒 to the π‘₯, both with respect to π‘₯. We know that the antiderivative of the derivative of a function is just equal to that function itself plus a constant of integration. So, this gives us 𝑒 to the π‘₯ plus the constant of integration is equal to the integral of 𝑒 to the π‘₯ with respect to π‘₯.

So, we can rewrite our integral of 𝑒 to the π‘₯ between one and zero as just 𝑒 to the π‘₯ evaluated between one and zero. Again, we don’t need to worry about our constant of integration since it will cancel when we evaluate the bounds of our definite integral. We’re now ready to evaluate each of these expressions between one and zero. Evaluating π‘₯ to the 𝑒 plus one divided by 𝑒 plus one between the limits of one and zero gives us one to the 𝑒 plus one over 𝑒 plus one minus zero to the 𝑒 plus one divided by 𝑒 plus one. And evaluating 𝑒 to the π‘₯ between our limits of one and zero gives us 𝑒 to the first power minus 𝑒 to zeroth power.

We know that zero to any positive power is just equal to zero, so we can remove the term zero to the 𝑒 plus one divided by 𝑒 plus one. We can simplify 𝑒 to the first power to just be equal to 𝑒. And we know that any nonzero number raised to the zeroth power is equal to one. So, we can simplify 𝑒 to the zeroth power to just be equal to one.

Finally, we know that one to the power of 𝑛 is just equal to one. So, we can replace our term of one to the 𝑒 plus one with just one. We can then expand, simplify, and rearrange to get that the integral of π‘₯ to the power of 𝑒 plus two multiplied by 𝑒 the π‘₯ with respect to π‘₯ is equal to negative two plus one divided by 𝑒 plus one plus two multiplied by 𝑒.

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