Basic Integration Problems
Basic Integration Problems

### Video Transcript

Evaluate the definite integral
between zero and one of the cube root of 𝑡 𝐢 plus one over 𝑡 plus one 𝐣 plus 𝑒
to the power of negative 𝑡 𝐤 with respect to 𝑡.

Remember, to integrate a
vector-valued function, we simply integrate each component in the usual way. And since this is a definite
integral, we can use the fundamental theorem of calculus and apply that to
continuous vector functions as shown. Here, capital 𝐑, of course, is the
antiderivative of lowercase 𝐫. So let’s integrate each component
with respect to 𝑡 and evaluate it between the limits of zero and one in the usual
way. We’ll begin by integrating the
component function for 𝐢. That’s the cube root of 𝑡 with
respect to 𝑡. And in fact, if we write that as 𝑡
to the power of one-third, it does become a little easier. We add one to the power and then
divide by that new value. So we get three-quarters 𝑡 to the
power of four over three. Then, evaluating this between the
limits of zero and one, we get three-quarters of one to the power of four over three
minus three-quarters of zero to the power of four over three, which is simply
three-quarters.

Next, we integrate the component
for 𝐣. That’s one over one plus 𝑡. We then recall that the indefinite
integral of 𝑓 prime of 𝑡 over 𝑓 of 𝑡 with respect to 𝑡 is equal to the natural
log of the absolute value of 𝑓 of 𝑡 plus some constant of integration 𝑐. Well, one is equal to the
derivative of one plus 𝑡. So the integral of one over one
plus 𝑡 is the natural log of the absolute value of one plus 𝑡. And we need to evaluate this
between zero and one. Notice how I’ve not included the
constant of integration here because this is a definite integral. And they cancel out. So we have the natural log of one
plus one minus the natural log of one plus zero. We no longer need the absolute
value sign because we know both one plus one and one plus zero are greater than
zero. So this is the natural log of two
minus the natural log of one, which is, of course, just the natural log of two.

Finally, we integrate our component
function for 𝐤. It’s the definite integral between
zero and one of 𝑒 to the power of negative 𝑡. The integral of 𝑒 to the power of
negative 𝑡 is negative 𝑒 to the power of negative 𝑡. So we have negative 𝑒 to the power
of negative one minus negative 𝑒 to the power of zero, which becomes negative 𝑒 to
the power of negative one plus one or one minus one over 𝑒. We’ve now successfully integrated
each of our component functions. So let’s put them back into vector
form. Our definite integral is equal to
three-quarters 𝐢 plus the natural log of two 𝐣 plus one minus one over 𝑒 𝐤. It’s important to realise that this
procedure can even be applied to initial value problems.

You are watching: Question Video: Finding the Value of an Unknown in a Polynomial Function given the Value of Its First Derivative at a Point. Info created by Bút Chì Xanh selection and synthesis along with other related topics.