Question Video: Evaluating a Vector-Valued Integral | Nagwa Question Video: Evaluating a Vector-Valued Integral | Nagwa

Question Video: Evaluating a Vector-Valued Integral Mathematics • Higher Education

Calculate the integral ∫[1/(𝑡 + 6) 𝐢 + 1/(2𝑡 + 5)⁴ 𝐣 + 5𝑡⁴/(𝑡⁵ + 5) 𝐤] d𝑡.

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Video Transcript

Calculate the integral of one over 𝑡 plus six 𝐢 plus one over two 𝑡 plus five all raised to the fourth power 𝐣 plus five 𝑡 to the fourth power divided by 𝑡 to the fifth power plus five 𝐤 with respect to 𝑡.

We’re asked to evaluate the integral of a vector-valued function. And to do this, we need to recall that we do this component-wise. In other words, we need to evaluate the integral of each component function separately. Let’s start by evaluating the integral of our first component function. That’s the integral of one divided by 𝑡 plus six with respect to 𝑡. There’s several different ways of evaluating this integral. For example, we could substitute 𝑢 is equal to 𝑡 plus six. And this would work. However, the easiest way is to recall the following integral result.

We know the integral of 𝑓 prime of 𝑡 divided by 𝑓 of 𝑡 with respect to 𝑡 is equal to the natural logarithm of the absolute value of 𝑓 of 𝑡 plus a constant of integration 𝑐. And we can see this is exactly what we have in our integral. The function 𝑓 of 𝑡 in our denominator would be 𝑡 plus six. And then we can see the derivative of 𝑡 plus six with respect to 𝑡 would be the coefficient of 𝑡, which is one, which is exactly equal to our numerator. So we can evaluate this integral by setting our function 𝑓 of 𝑡 to be 𝑡 plus six. This gives us the natural logarithm of the absolute value of 𝑡 plus six plus a constant of integration we’ll call lowercase 𝑎.

We now want to evaluate the integral of our second component function. That means we want to evaluate the integral of one divided by two 𝑡 plus five all raised to the fourth power with respect to 𝑡. This time, we’ll do this by using a substitution. We’re going to use the substitution 𝑢 is equal to two 𝑡 plus five. We then need to differentiate both sides of this expression with respect to 𝑡. We get d𝑢 by d𝑡 will be the derivative of two 𝑡 plus five with respect to 𝑡. This is a linear function. So its derivative with respect to 𝑡 will be the coefficient of 𝑡, which is two.

Now although we know d𝑢 by d𝑡 is not a fraction, we can treat it a little bit like a fraction when we’re doing integration by substitution. This gives us the equivalent statement in terms of differentials: d𝑢 is equal to two d𝑡. And one step that can be helpful is to notice in our integral we don’t have the expression two d𝑡. Instead, we have one d𝑡. So we’ll divide both sides of our equation in terms of differentials by two. This gives us one-half d𝑢 is equal to one d𝑡.

We’re now ready to evaluate our integral by substitution. First, we use our substitution 𝑢 is equal to two 𝑡 plus five to rewrite our denominator as 𝑢 to the fourth power. Then, all we need to do is use our equation in terms of differentials to rewrite our integral as the integral of one-half over 𝑢 to the fourth power with respect to 𝑢. And we can simplify this integral. First, we’ll take the constant factor of one-half outside of our integral.

Next, by using our laws of exponents, we’ll rewrite one over 𝑢 to the fourth power as 𝑢 to the power of negative four. This means we need to evaluate the integral of one-half times 𝑢 to the power of negative four with respect to 𝑢. We can do this by using the power rule for integration. We want to add one to our exponent of 𝑢, giving us a new exponent of negative three, and then divide by this new exponent. This gives us 𝑢 to the power of negative three divided by two times negative three. And of course, we need to add a constant of integration. We’ll call our constant of integration lowercase 𝑏.

And we can simplify this. In our denominator, two multiplied by negative three is equal to negative six. And instead of multiplying by 𝑢 to the power of negative three, we can instead divide by 𝑢 cubed. This gives us negative one over six 𝑢 cubed plus 𝑏. But remember, we want to give our expression in terms of 𝑡. To do this, we’re going to use the substitution 𝑢 is equal to two 𝑡 plus five. Doing this, we get the integral of our second component function with respect to 𝑡 is equal to negative one divided by six times two 𝑡 plus five all cubed plus 𝑏.

We now want to find the integral of our third component function. We’ll start by clearing some space. We now want to find the integral of our third component function. That’s the integral of five 𝑡 to the fourth power divided by 𝑡 to the fifth power plus five with respect to 𝑡. Once again, we can do this by substitution. We could use the substitution 𝑢 is equal to 𝑡 to the fifth power plus five. And the reason we might choose this substitution is the derivative of this expression appears in our numerator.

And this would work. However, we’ve already done this in the general case, and we’ve written our formula earlier. In this case, our function 𝑓 of 𝑡 would be 𝑡 to the fifth power plus five. And then by using the power rule for differentiation, we know that 𝑓 prime of 𝑡 would be five 𝑡 to the fourth power, which is our numerator. So we can just integrate this expression with respect to 𝑡 by setting 𝑓 of 𝑡 to be 𝑡 to the fifth power plus five. This gives us the natural logarithm of the absolute value of 𝑡 to the fifth power plus five plus a constant of integration we’ll call lowercase 𝑐.

Now that we’ve integrated all three of the component functions with respect to 𝑡, we’re ready to evaluate the integral of the vector-valued function given to us in the question. First, we showed the integral of the first component function is the natural logarithm of the absolute value of 𝑡 plus six plus 𝑎. Next, we showed the integral of our second component function is negative one divided by six times two 𝑡 plus five all cubed plus 𝑏. And it’s worth pointing out here we took the negative one outside of our vector and used the fact that lowercase 𝑏 is a constant of integration. Finally, we were able to show the integral of our third component function was the natural logarithm of the absolute value of 𝑡 to the fifth power plus five plus our constant of integration lowercase 𝑐.

So when we integrate our vector-valued function, we just write the integrals as the new component functions. This gives us the following expression for our vector-valued function. And we could leave our answer like this. However, there’s something worth noticing. If we were to distribute our parentheses, we would notice we would get the term 𝑎𝐢 plus 𝑏𝐣 plus 𝑐𝐤. We know that lowercase 𝑎, 𝑏, and 𝑐 are all constants of integration. So 𝑎𝐢, 𝑏𝐣, and 𝑐𝐤 would just be constant vectors. So we could combine all of these into a new vector. We’ll combine these into a new constant vector we’ll call capital 𝐂. And this gives us our final answer.

Therefore, we were able to show the integral of one over 𝑡 plus six 𝐢 plus one over two 𝑡 plus five all raised to the fourth power 𝐣 plus five 𝑡 to the fourth power over 𝑡 to the fifth power plus five 𝐤 with respect to 𝑡 is equal to the natural logarithm of the absolute value of 𝑡 plus six 𝐢 minus one over six times two 𝑡 plus five all cubed 𝐣 plus the natural logarithm of the absolute value of 𝑡 to the fifth power plus five 𝐤 plus a constant vector 𝐂.

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