### Video Transcript

Evaluate the integral the integral
from zero to two of four π‘ plus three π’ plus π‘ cubed plus three π‘ π£ with
respect to π‘.

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. Weβll start by finding the definite
integral of the first component function. Thatβs the integral from zero to
two of four π‘ plus three with respect to π‘. We can evaluate this integral by
using the power rule for integration. Which tells us for constants π and
π, where π is not equal to negative one, to integrate π times π‘ to the πth
power with respect to π‘, we add one to the exponent and then divide by this new
exponent. Giving us ππ‘ to the power of π
plus one divided by π plus one plus a constant of integration π.

By writing four π‘ as four π‘ to
the first power and three as three π‘ to the zeroth power, we can integrate this
using the power rule for integration. We get four π‘ squared over two
plus three π‘ to the first power over one plus a constant of integration π. 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 π‘ to the first power over
one is just equal to three π‘. And four over two is just equal to
two.

So we now need to evaluate two π‘
squared plus three π‘ at the limits of our integral. π‘ is equal to zero and π‘ 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 π‘ cubed plus three π‘ with respect to π‘. Again, we evaluate this by using
the power rule for integration. This gives us π‘ to the fourth
power over four plus three π‘ squared over two evaluated at the limits of our
integral, π‘ is equal to zero and π‘ 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 itβs equal to 10. Since weβve now evaluated the
definite integral of both of our component functions, weβve shown the integral from
zero to two of four π‘ plus three π’ plus π‘ cubed plus three π‘ π£ with respect to
π‘ is equal to 14π’ plus 10π£. 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.