Suppose that 𝐹 of 𝑡 is equal to
𝑡 squared. Evaluate the integral between one
and three of 𝐹 prime of 𝑡 with respect to 𝑡 using the net change theorem.
For this question, and many others,
it’s always helpful to write out the theorem that you’ll be using. Here, we have the net change
theorem. Although the variable in our
question is 𝑡 and the variable in the theorem we’ve written out is 𝑥, things apply
in exactly the same way. We take the lower limit of
integration, or 𝑎, as one and the upper limit of integration, or 𝑏, as three. Using our theorem, we’re able to
say that the integral between one and three of 𝐹 prime of 𝑡 with respect to 𝑡 is
equal to 𝐹 of three minus 𝐹 of one.
Again, here’re the limits of
integration that we’ve used. Notice, that we’re able to apply
our theorem because we have a definite integral of a derivative 𝐹 prime of 𝑡. And this can be thought of as the
rate of change of 𝐹 of 𝑡. Now, the function itself 𝐹 of 𝑡
has been given in the question. And it’s simply 𝑡 squared. We can, therefore, evaluate the
right-hand side of our function as three squared minus one squared. Of course, this is nine minus one,
which is simply equal to eight. Following this simplification, we
have answered our question. We have used the net change theorem
to evaluate that the given integral is equal to eight.