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, weβve 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 weβll see why in a moment. To move forward with this question,
weβre 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 π, must be continuous on the closed interval
between π and π. π and π are the limits of
integration, which in our case are four and nine. Now, the function that weβre now
working with, lowercase π, is the square root of π₯, which weβve 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 itβs 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, weβre 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 π was continuous over the closed interval
between the limits of integration. One final point that we didnβt
really go into earlier. Weβre 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.
