Video: Finding the Value of an Unknown in a Polynomial Function given the Value of Its First Derivative at a Point

>Evaluate the integral ∫_(0)^(1) (βˆ›π‘‘ 𝐒 + (1/(𝑑 + 1) 𝐣) + 𝑒^(βˆ’π‘‘) 𝐀) d𝑑.

02:30

Video Transcript

Evaluate the definite integral between zero and one of the cube root of 𝑑 𝐒 plus one over 𝑑 plus one 𝐣 plus 𝑒 to the power of negative 𝑑 𝐀 with respect to 𝑑.

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 𝑑 and evaluate it between the limits of zero and one in the usual way. We’ll begin by integrating the component function for 𝐒. That’s the cube root of 𝑑 with respect to 𝑑. And in fact, if we write that as 𝑑 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 𝑑 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 𝑑. We then recall that the indefinite integral of 𝑓 prime of 𝑑 over 𝑓 of 𝑑 with respect to 𝑑 is equal to the natural log of the absolute value of 𝑓 of 𝑑 plus some constant of integration 𝑐. Well, one is equal to the derivative of one plus 𝑑. So the integral of one over one plus 𝑑 is the natural log of the absolute value of one plus 𝑑. 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 𝐀. It’s the definite integral between zero and one of 𝑒 to the power of negative 𝑑. The integral of 𝑒 to the power of negative 𝑑 is negative 𝑒 to the power of negative 𝑑. 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 𝐒 plus the natural log of two 𝐣 plus one minus one over 𝑒 𝐀. It’s important to realise that this procedure can even be applied to initial value problems.

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