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

Evaluate the integral ∫_0^2 (𝑑 𝑖 + 𝑑³ 𝑗 + 3𝑑⁡ π‘˜) d𝑑.

01:32

Video Transcript

Evaluate the definite integral between zero and two of 𝑑𝑖 plus 𝑑 cubed 𝑗 plus three 𝑑 to the fifth power π‘˜, with respect to 𝑑.

In this question, we see that we’re looking to find a definite integral of a vector-valued function. This might look a little scary, but all we do is integrate each component in turn. So, we’ll evaluate the definite integral between zero and two of 𝑑 with respect to 𝑑. We’ll repeat the process for 𝑑 cubed, and finally for the component for π‘˜. That’s three 𝑑 to the fifth power.

We then recall that to integrate polynomial terms whose exponent is not equal to negative one, we add one to the exponent and then divide by that new value. So, the integral of 𝑑 is 𝑑 squared over two. We need to evaluate this between the limits of zero and two. So, that’s two squared over two minus zero squared over two, which is simply two.

In the same way, the integral of 𝑑 cubed is 𝑑 to the fourth power over four. Between the limits of zero and two, this becomes two to the fourth power over four minus zero to the fourth power over four, which is four.

Then finally, the integral of three 𝑑 to the fifth power is three 𝑑 to the sixth power over six, which simplifies to 𝑑 to the sixth power over two. Evaluating this between our limits and we get two to the sixth power over two minus zero to the sixth power over two, which is 32. And we can see that the component of our definite integral for 𝑖 is two. For 𝑗, it’s four. And for π‘˜, it’s 32. Our definite integral is two 𝑖 plus four 𝑗 plus 32π‘˜.

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