### 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.