Math / Calculus

Question $\left|\mid \lim _{-0} \sum_{k=1}^{n} \sqrt{c^{2}+9} \Delta x_{k}\right.$ where $P$ is a partition of $[-6,2]$

Studdy Solution

STEP 1

What is this asking? We need to find the definite integral of the square root of $c^2 + 9$ from $-6$ to $2$ . Watch out! Don't forget that $c^2 + 9$ is a constant, so the integral is super easy!

STEP 2

1. Rewrite the limit as a definite integral.
2. Calculate the definite integral.

STEP 3

Alright, let's rewrite this limit of a sum as a definite integral!
Remember, that's exactly what the limit represents.
We've got $\lim_{\|P\| \to 0} \sum_{k=1}^{n} \sqrt{c^{2}+9} \Delta x_{k}$ , where $P$ is a partition of $[-6, 2]$ .

STEP 4

This translates directly to the definite integral: $\int_{-6}^{2} \sqrt{c^2 + 9} \, dx$ See how clean that is?
The function we're integrating is $\sqrt{c^2 + 9}$ , and we're integrating it with respect to $x$ from $-6$ to $2$ .

STEP 5

Now, the fun part!
Let's calculate this integral.
Notice that $\sqrt{c^2 + 9}$ is a constant with respect to $x$ .
That makes our lives much easier!

STEP 6

When we integrate a constant, we just multiply it by $x$ .
So, we have: $\int \sqrt{c^2 + 9} \, dx = \sqrt{c^2 + 9} \cdot x$

STEP 7

Now, we need to evaluate this from $-6$ to $2$ .
That means we plug in $2$ for $x$ , then subtract what we get when we plug in $-6$ for $x$ .

STEP 8

Plugging in $2$ , we get: $\sqrt{c^2 + 9} \cdot 2 = 2\sqrt{c^2 + 9}$

STEP 9

Plugging in $-6$ , we get: $\sqrt{c^2 + 9} \cdot (-6) = -6\sqrt{c^2 + 9}$

STEP 10

Now, subtract: $2\sqrt{c^2 + 9} - (-6\sqrt{c^2 + 9}) = 2\sqrt{c^2 + 9} + 6\sqrt{c^2 + 9} = 8\sqrt{c^2 + 9}$

STEP 11

So, our final answer is $8\sqrt{c^2 + 9}$ !

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