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Find the average value of the function on the given interval.

$ g(t) = \dfrac{t}{\sqrt{3 + t^2}} $ , $ [1, 3] $

$\sqrt{3}-1 \approx 0.732$

Calculus 2 / BC

Chapter 6

Applications of Integration

Section 5

Average Value of a Function

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we know we're using the average value formula, which is one over B, which is three minus a, which is one from the bounds of 80 B or 123 or function T over three plus t squared square rooted D T. We know we can simplify this to be 1/4 times two and then we know we should actually do U substitution over here. So let's define our you is three plus t squared. Therefore, our do your derivative is to t d t right Because the three cancels, it's just a constant. Therefore, we have two year the 1/2 from 4 to 12 plugging in 1/2 time squirt of 12 minus squared of four gives us sward of three minus one. This is also known as 0.732 if you prefer in the decibel form.

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