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Calculate the integral of π‘π’ plus three π£ with respect to π‘.
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Weβre given an indefinite integral which weβre asked to evaluate.
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And we can see something interesting about our integrand, In this case, weβre given a vector-valued function.
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And we know this is a vector-valued function because it contains the unit directional vectors π’ and π£.
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And itβs also worth pointing out here thereβs a lot of different notation for vectors.
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For example, we can see the hat notation, which represents unit vectors.
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However, you often also see vectors written in bold notation, underlined with full arrows or half-arrows.
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It doesnβt matter which notation you prefer; they all mean the same thing.
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So now, we need to recall how to integrate a vector-valued function.
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To do this, we just need to evaluate the integral of each of our component functions.
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So, weβll start with evaluating the integral of our first component function.
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Thatβs the integral of π‘ with respect to π‘.
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And we can evaluate this integral by using the power rule for integration.
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We recall this tells us for all constants π not equal to negative one, the integral of π‘ to the πth power with respect to π‘ is equal to π‘ to the power of π plus one divided by π plus one plus the constant of integration π.
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In our case, we can write π‘ as π‘ to the first power.
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So, our value of π is equal to one.
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So, we add one to our exponent of one, giving us two, and then divide by this new exponent.
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This gives us π‘ squared over two.
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And weβll add a constant of integration weβll call π.
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Next, we need to evaluate the integral of our second component function.
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Thatβs the integral of three with respect to π‘.
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Thereβs a few different ways of doing this.
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For example, we could rewrite three as three times π‘ to the zeroth power and then use the power rule for integration.
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If we did this, we would get three π‘ to the first power divided by one plus the constant of integration π.
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But of course, dividing by one and raising a number to the first power doesnβt change this value.
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So, we just get three π‘ plus π.
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Weβre now ready to use these to evaluate the integral of our vector-valued function.
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That is the integral of π‘π’ plus three π£ with respect to π‘.
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Now, all we need to do is integrate this component wise.
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And we found the integral of our first component is π‘ squared over two plus π and the integral of our second component is three π‘ plus π.
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So, we get the integral of our vector-valued function with respect to π‘ is π‘ squared over two plus π times π’ plus three π‘ plus π times π£.
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And we could leave our answer like this.
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However, thereβs one more piece of simplification we could do.
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If we were to distribute our parentheses, we see we would end up with π times π’ plus π times π£.
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But remember, π and π are constants of integration.
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So, this is just equal to a constant vector.
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So, we can just call this constant vector the vector π.
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This gives us π‘ squared over two π’ plus three π‘π£ plus our constant vector π.
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And this is our final answer.
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Therefore, by evaluating the integral of the vector-valued function π‘π’ plus three π£ with respect to π‘ componentwise, we were able to show that this must be equal to π‘ squared over two π’ plus three π‘π£ plus π.