Video: Evaluating the Definite Integral of a Vector-Valued Function

Evaluate the integral โˆซ_(1) ^(4) (2๐‘ก^(3/2) ๐ข + (๐‘ก + 1)โˆš(๐‘ก) ๐ค) d๐‘ก.

06:33

Video Transcript

Evaluate the integral from one to four of two ๐‘ก to the power of three over two ๐ข plus ๐‘ก plus one multiplied by the square root of ๐‘ก ๐ค with respect to ๐‘ก.

We can see weโ€™re asked to evaluate a definite integral, and we can see that this contains the vectors ๐ข and ๐ค. And we know these are the unit directional vectors. We know that theyโ€™re unit vectors because they have the hat notation. And in fact, thereโ€™s a lot of different notations for vectors. For example, we often use bold letters, underlined letters, or half-arrow notation. All of these notations mean roughly the same thing. They represent vectors.

So in this question, weโ€™re being asked to evaluate the definite integral of a vector-valued function. And to find the integral of a vector-valued function, we do this component-wise. However, thereโ€™s one small problem. We can see that we have our vectors ๐ข and ๐ค. But we donโ€™t have our vector ๐ฃ. But itโ€™s fine because we can just add this in as zero ๐ฃ. Now, we want to evaluate the definite integral from four to one component-wise.

Letโ€™s start with our first component function. Thatโ€™s two ๐‘ก to the power of three over two. We want to calculate the definite integral from one to four of two ๐‘ก to the power of three over two with respect to ๐‘ก. And we know we can do this by using the power rule for integration. We recall this tells us for any real constants ๐‘Ž and ๐‘› where ๐‘› is not equal to negative one, the integral of ๐‘Ž times ๐‘ก to the ๐‘›th power with respect to ๐‘ก is equal to ๐‘Ž times ๐‘ก to the power of ๐‘› plus one divided by ๐‘› plus one plus a constant of integration ๐ถ. We add one to our exponent and then divide by this new exponent.

Of course, in this case, weโ€™re evaluating a definite integral, so we donโ€™t need our constant of integration because it will cancel in the working. So we add one to our exponent of three over two giving us five over two, and then we divide by five over two. This gives us two times ๐‘ก to the power of five over two divided by five over two evaluated at limits of integration one and four. And instead of dividing by five over two, we can instead multiply it by the reciprocal of five over two. This gives us two times ๐‘ก to the power of five over two multiplied by two over five evaluated at the limits of integration one and four.

And we can then simplify this. Two multiplied by two gives us four. So weโ€™ve simplified this to give us four-fifths times ๐‘ก to the power of five over two evaluated at limits of integration one and four. All we need to do now is evaluate this at the limits of integration. Evaluating this at the limits of integration, we get four-fifths times four raised to the power of five over two minus four-fifths times one raised to the power of five over two.

To evaluate this expression, recall raising a number to the power of five over two is the same as raising it to the power of one-half and then raising it to the power of five. But raising a number to the power of one-half is the same as taking the square root, so we can rewrite four to the power of five over two as the square root of four all raised to the fifth power. And of course, the square root of four is two, so we get two to the fifth power, which we can evaluate is equal to 32. So our first term simplifies to give us four-fifths times 32.

To simplify our second term, we need to recall that one raised to any power just simplifies to give us one. So our second term simplifies to give us four over five. And we now have four over five times 32 minus four over five. And if we evaluate this expression, weโ€™ll get 124 divided by five. So we evaluated the definite integral from one to four of our first component function and got 124 divided by five.

We now need to do the same with our second component function. Remember that our second component function is zero, so this gives us the integral from one to four of zero with respect to ๐‘ก. Now we could evaluate this integral directly. However, we also know the definite integral of zero is always equal to zero, so weโ€™ll just write this as zero. And now weโ€™ve calculated the definite integral of our second component function. We got zero.

We now need to calculate the definite integral of our third component function. Thatโ€™s the integral from one to four of ๐‘ก plus one multiplied by root ๐‘ก with respect to ๐‘ก. And to do this, weโ€™ll start by using our laws of exponents to rewrite root ๐‘ก as ๐‘ก to the power of one-half. Now, if we distribute ๐‘ก to the power of one-half over our parentheses, weโ€™ll be able to evaluate this integral by using the power rule for integration. Distributing this over our parentheses, we get the integral from one to four of ๐‘ก times ๐‘ก to the power of one-half plus ๐‘ก to the power of one-half with respect to ๐‘ก.

Remember, we can rewrite ๐‘ก as ๐‘ก to the first power. Then remember, by using our laws of exponents, ๐‘ก to the first power multiplied by ๐‘ก to the power of one-half will be ๐‘ก to the power of one plus one-half, which is ๐‘ก to the power of three over two. So now we need to find the definite integral from one to four of ๐‘ก to the power of three over two plus ๐‘ก to the power of one-half with respect to ๐‘ก. Of course, we can do this by using the power rule for integration. We just need to add one to our exponent of ๐‘ก and then divide by this exponent of ๐‘ก. This gives us the following expression.

And we know we can simplify this. In our first term, instead of dividing by five over two, weโ€™ll multiply by the reciprocal of five over two. And in our second term, instead of dividing by three over two, weโ€™ll multiply by the reciprocal of three over two. And if we do this, we get two times ๐‘ก to the power of five over two divided by five plus two times ๐‘ก to the power of three over two divided by three evaluated at the limits of integration one and four.

Now, all we need to do is evaluate this at the limits of integration. Doing this gives us two times four to the power of five over two divided by five plus two times four to the power of three over two divided by three minus two times one to the power of five over two divided by five plus two times one to the power of three over two divided by three. And we could evaluate this as we did before by using our laws of exponents. Or we could just use a calculator. If we do this, we get 256 divided by 15.

And now weโ€™re ready to evaluate the definite integral of the vector-valued function given to us in the question. First, we found the definite integral of our first component function was 124 divided by five. So our answer will have 124 divided by five ๐ข. Next, when we found the definite integral of our second component function, we got zero. So our answer will have zero lots of ๐ฃ. Of course, we donโ€™t need to include this since its coefficient is zero. Finally, when we evaluated the definite integral of our third component function, we got 256 divided by 15. So our answer will have 256 divided by 15 ๐ค. And this is our final answer.

In this video, we were asked to evaluate the definite integral of a vector-valued function, and we did this component-wise. This allowed us to get the answer 124 divided by five ๐ข plus 256 over 15 ๐ค.

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