Question Video: Calculating a Definite Integral of a Vector-Valued Function Mathematics

Video Transcript

Evaluate the integral the integral from zero to two of four 𝑡 plus three 𝐢 plus 𝑡 cubed plus three 𝑡 𝐣 with respect to 𝑡.

The question is asking us to evaluate a definite integral of a vector-valued function. To do this, we just find the definite integral of each of the component functions. We’ll start by finding the definite integral of the first component function. That’s the integral from zero to two of four 𝑡 plus three with respect to 𝑡. We can evaluate this integral by using the power rule for integration. Which tells us for constants 𝑎 and 𝑛, where 𝑛 is not equal to negative one, to integrate 𝑎 times 𝑡 to the 𝑛th power with respect to 𝑡, we add one to the exponent and then divide by this new exponent. Giving us 𝑎𝑡 to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant of integration 𝑐.

By writing four 𝑡 as four 𝑡 to the first power and three as three 𝑡 to the zeroth power, we can integrate this using the power rule for integration. We get four 𝑡 squared over two plus three 𝑡 to the first power over one plus a constant of integration 𝑐. However, this is a definite integral. So we need to evaluate this at the limits of our integral. And since this is a definite integral, when we evaluate this at the limits of our integral, the constants of integration will cancel. We can also simplify this. Three 𝑡 to the first power over one is just equal to three 𝑡. And four over two is just equal to two.

So we now need to evaluate two 𝑡 squared plus three 𝑡 at the limits of our integral. 𝑡 is equal to zero and 𝑡 is equal to two. Evaluating this at the limits of our integral gives us two times two squared plus three times two minus two times zero squared plus three times zero, which we can calculate to give us 14. We then want to do the same to our second component function. We need to calculate the integral from zero to two of 𝑡 cubed plus three 𝑡 with respect to 𝑡. Again, we evaluate this by using the power rule for integration. This gives us 𝑡 to the fourth power over four plus three 𝑡 squared over two evaluated at the limits of our integral, 𝑡 is equal to zero and 𝑡 is equal to two.

Evaluating this at the limits of our integral gives us two to the fourth power over four plus three times two squared over two minus zero to the fourth power over four plus three times zero squared over two. And when we calculate this expression, we see it’s equal to 10. Since we’ve now evaluated the definite integral of both of our component functions, we’ve shown the integral from zero to two of four 𝑡 plus three 𝐢 plus 𝑡 cubed plus three 𝑡 𝐣 with respect to 𝑡 is equal to 14𝐢 plus 10𝐣. And this is because the integral from zero to two of our first component function was equal to 14. And the integral of our second component function from zero to two was equal to 10.