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 β«[1/(π‘ + 6) π’ + 1/(2π‘ + 5)β΄ π£ + 5π‘β΄/(π‘β΅ + 5) π€] dπ‘.

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### Video Transcript

Calculate the integral of one over π‘ plus six π’ plus one over two π‘ plus five all raised to the fourth power π£ plus five π‘ to the fourth power divided by π‘ to the fifth power plus five π€ with respect to π‘.

Weβ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. Letβs start by evaluating the integral of our first component function. Thatβs the integral of one divided by π‘ plus six with respect to π‘. Thereβs several different ways of evaluating this integral. For example, we could substitute π’ is equal to π‘ plus six. And this would work. However, the easiest way is to recall the following integral result.

We know the integral of π prime of π‘ divided by π of π‘ with respect to π‘ is equal to the natural logarithm of the absolute value of π of π‘ plus a constant of integration π. And we can see this is exactly what we have in our integral. The function π of π‘ in our denominator would be π‘ plus six. And then we can see the derivative of π‘ plus six with respect to π‘ would be the coefficient of π‘, which is one, which is exactly equal to our numerator. So we can evaluate this integral by setting our function π of π‘ to be π‘ plus six. This gives us the natural logarithm of the absolute value of π‘ plus six plus a constant of integration weβll call lowercase π.

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 π‘ plus five all raised to the fourth power with respect to π‘. This time, weβll do this by using a substitution. Weβre going to use the substitution π’ is equal to two π‘ plus five. We then need to differentiate both sides of this expression with respect to π‘. We get dπ’ by dπ‘ will be the derivative of two π‘ plus five with respect to π‘. This is a linear function. So its derivative with respect to π‘ will be the coefficient of π‘, which is two.

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

Weβre now ready to evaluate our integral by substitution. First, we use our substitution π’ is equal to two π‘ plus five to rewrite our denominator as π’ 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 π’ to the fourth power with respect to π’. And we can simplify this integral. First, weβll take the constant factor of one-half outside of our integral.

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

And we can simplify this. In our denominator, two multiplied by negative three is equal to negative six. And instead of multiplying by π’ to the power of negative three, we can instead divide by π’ cubed. This gives us negative one over six π’ cubed plus π. But remember, we want to give our expression in terms of π‘. To do this, weβre going to use the substitution π’ is equal to two π‘ plus five. Doing this, we get the integral of our second component function with respect to π‘ is equal to negative one divided by six times two π‘ plus five all cubed plus π.

We now want to find the integral of our third component function. Weβll start by clearing some space. We now want to find the integral of our third component function. Thatβs the integral of five π‘ to the fourth power divided by π‘ to the fifth power plus five with respect to π‘. Once again, we can do this by substitution. We could use the substitution π’ is equal to π‘ 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, weβve already done this in the general case, and weβve written our formula earlier. In this case, our function π of π‘ would be π‘ to the fifth power plus five. And then by using the power rule for differentiation, we know that π prime of π‘ would be five π‘ to the fourth power, which is our numerator. So we can just integrate this expression with respect to π‘ by setting π of π‘ to be π‘ to the fifth power plus five. This gives us the natural logarithm of the absolute value of π‘ to the fifth power plus five plus a constant of integration weβll call lowercase π.

Now that weβve integrated all three of the component functions with respect to π‘, weβ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 π‘ plus six plus π. Next, we showed the integral of our second component function is negative one divided by six times two π‘ plus five all cubed plus π. And itβs worth pointing out here we took the negative one outside of our vector and used the fact that lowercase π 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 π‘ to the fifth power plus five plus our constant of integration lowercase π.

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, thereβs something worth noticing. If we were to distribute our parentheses, we would notice we would get the term ππ’ plus ππ£ plus ππ€. We know that lowercase π, π, and π are all constants of integration. So ππ’, ππ£, and ππ€ would just be constant vectors. So we could combine all of these into a new vector. Weβll combine these into a new constant vector weβll call capital π. And this gives us our final answer.

Therefore, we were able to show the integral of one over π‘ plus six π’ plus one over two π‘ plus five all raised to the fourth power π£ plus five π‘ to the fourth power over π‘ to the fifth power plus five π€ with respect to π‘ is equal to the natural logarithm of the absolute value of π‘ plus six π’ minus one over six times two π‘ plus five all cubed π£ plus the natural logarithm of the absolute value of π‘ to the fifth power plus five π€ plus a constant vector π.

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