Video: Evaluating the Definite Integral of a Root Function

Evaluate ∫_(4) ^(9) βˆ’2√π‘₯ dπ‘₯.

02:14

Video Transcript

Evaluate the integral between four and nine of negative two times the square root of π‘₯ with respect to π‘₯.

For this question, we’ve been asked to evaluate a definite integral. With questions of this type, it can sometimes be useful to move constant factors, such as the negative two, from the inside of the integrand to the outside. Next, we might also find it useful to reexpress our square root of π‘₯ as π‘₯ to the power of a half or π‘₯ to the power of 0.5. And we’ll see why in a moment. To move forward with this question, we’re gonna be using the second part of the fundamental theorem of calculus. This gives us a way to evaluate definite integrals using the antiderivative of the function which forms the integrand.

At this point, we note that the theorem states the function, lowercase 𝑓, must be continuous on the closed interval between π‘Ž and 𝑏. π‘Ž and 𝑏 are the limits of integration, which in our case are four and nine. Now, the function that we’re now working with, lowercase 𝑓, is the square root of π‘₯, which we’ve just expressed as π‘₯ to the power of a half. This function is not continuous over the entire set of real numbers, but rather is only continuous when π‘₯ is greater than or equal to zero. Luckily, both of the limits of our definite integral, four and nine, are greater than or equal to zero. And so we can therefore say that the square root of π‘₯ is continuous on the closed interval between four and nine. This means that it’s perfectly fine to use our theorem.

To proceed with our evaluation, we use the power rule of integration, raising the power of π‘₯ by one and dividing by the new power. The antiderivative of π‘₯ to the power of half is therefore two over three times π‘₯ to the power of three over two. Again, we’re gonna shift this constant factor outside of our brackets to make our calculations easier. Next ,we substitute in our limits of integration. Now, at this stage, it might be more useful to see our power of three over two expressed as the cube of a square root. Conveniently, nine and four are both square numbers. And we can simplify our parentheses to be three cubed minus two cubed. We now move forward with a few more simplifications. And eventually, we reach an answer of negative 76 over three. This is the final answer to our question.

We evaluated the given definite integral using the second part of the fundamental theorem of calculus to help us. Along the way, we were sure to confirm that our function lowercase 𝑓 was continuous over the closed interval between the limits of integration. One final point that we didn’t really go into earlier. We’re able to say that the square root of π‘₯ is not continuous when π‘₯ is less than zero, because, in fact, the square root of π‘₯ is undefined over the real numbers when π‘₯ is less than zero. And of course, a function cannot be continuous at points where it is not defined.

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