Video Transcript
True or false, the net change
theorem states that the integral from zero to two of π to the power of two π₯ with
respect to π₯ is equal to a half multiplied by π to the fourth power minus one.
The question wants us to use the
net change theorem. What does the net change theorem
tell us? The net change theorem tells us
that the definite integral from π to π of a rate-of-change function π prime of π₯
with respect to π₯ is just equal to π evaluated at π minus π evaluated at π. Itβs the net change of our function
π. The question wants us to use the
net change theorem to evaluate the indefinite integral from zero to two of π to the
power of two π₯ with respect to π₯. So, weβll set π prime of π₯ equal
to π to the power of two π₯. And since our integral is from zero
to two, weβll set the lower limit of our integral, π, equal to zero and the upper
limit of our integral, π, equal to two.
Now, all we need to apply the net
change theorem is to find our function π of π₯. We can see that π of π₯ is an
antiderivative of our derivative function π prime of π₯. And we can find this antiderivative
by using our integration rule. We know, for any constant π, the
integral of π raised to the power of ππ₯ with respect to π₯ is equal to π to the
power of ππ₯ divided by π plus a constant of integration π. In our case, π is equal to
two. This means our function π of π₯ is
equal to π to the power of two π₯ divided by two plus a constant of integration
π. However, since weβre using a
definite integral, the constants of integration will cancel during our working. So, we donβt actually need to worry
about this.
So, by using the net change
theorem, we have the definite integral from zero to two of π to the power of two π₯
with respect to π₯ is equal to π of two minus π of zero. This gives us π to the power of
two times two divided by two minus π to the power of two times zero over two. And this is the point we see if we
have included our constant of integration, we wouldβve added π and subtracted
π. So, the constant of integration
wouldβve canceled. We can simplify this to give us π
to the fourth power over two minus a half since π to zeroth power is just equal to
one. Finally, we take out a factor of a
half to give a half multiplied by π to the fourth power minus one.
Therefore, weβve shown, by using
the net change theorem, the integral from zero to two of π to the power of two π₯
with respect to π₯ is equal to a half multiplied by π to the fourth power minus
one. Our statement was true.