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ipn [44]
3 years ago
15

How to find average value of a function over a given interval?

Mathematics
1 answer:
Strike441 [17]3 years ago
3 0
<span><span>f<span>(x)</span>=8x−6</span><span>f<span>(x)</span>=8x-6</span></span> , <span><span>[0,3]</span><span>[0,3]

</span></span>The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.<span><span>(−∞,∞)</span><span>(-∞,∞)</span></span><span><span>{x|x∈R}</span><span>{x|x∈ℝ}</span></span><span><span>f<span>(x)</span></span><span>f<span>(x)</span></span></span> is continuous on <span><span>[0,3]</span><span>[0,3]</span></span>.<span><span>f<span>(x)</span></span><span>f<span>(x)</span></span></span> is continuousThe average value of function <span>ff</span> over the interval <span><span>[a,b]</span><span>[a,b]</span></span> is defined as <span><span>A<span>(x)</span>=<span>1<span>b−a</span></span><span>∫<span>ba</span></span>f<span>(x)</span>dx</span><span>A<span>(x)</span>=<span>1<span>b-a</span></span><span>∫ab</span>f<span>(x)</span>dx</span></span>.<span><span>A<span>(x)</span>=<span>1<span>b−a</span></span><span>∫<span>ba</span></span>f<span>(x)</span>dx</span><span>A<span>(x)</span>=<span>1<span>b-a</span></span><span>∫ab</span>f<span>(x)</span>dx</span></span>Substitute the actual values into the formula for the average value of a function.<span><span>A<span>(x)</span>=<span>1<span>3−0</span></span><span>(<span>∫<span>30</span></span>8x−6dx)</span></span><span>A<span>(x)</span>=<span>1<span>3-0</span></span><span>(<span>∫03</span>8x-6dx)</span></span></span>Since integration is linear, the integral of <span><span>8x−6</span><span>8x-6</span></span> with respect to <span>xx</span> is <span><span><span>∫<span>30</span></span>8xdx+<span>∫<span>30</span></span>−6dx</span><span><span>∫03</span>8xdx+<span>∫03</span>-6dx</span></span>.<span><span>A<span>(x)</span>=<span>1<span>3−0</span></span><span>(<span>∫<span>30</span></span>8xdx+<span>∫<span>30</span></span>−6dx)</span></span><span>A<span>(x)</span>=<span>1<span>3-0</span></span><span>(<span>∫03</span>8xdx+<span>∫03</span>-6dx)</span></span></span>Since <span>88</span> is constant with respect to <span>xx</span>, the integral of <span><span>8x</span><span>8x</span></span> with respect to <span>xx</span> is <span><span>8<span>∫<span>30</span></span>xdx</span><span>8<span>∫03</span>xdx</span></span>.<span><span>A<span>(x)</span>=<span>1<span>3−0</span></span><span>(8<span>∫<span>30</span></span>xdx+<span>∫<span>30</span></span>−6dx)</span></span><span>A<span>(x)</span>=<span>1<span>3-0</span></span><span>(8<span>∫03</span>xdx+<span>∫03</span>-6dx)</span></span></span>By the Power Rule, the integral of <span>xx</span> with respect to <span>xx</span> is <span><span><span>12</span><span>x2</span></span><span><span>12</span><span>x2</span></span></span>.<span>A<span>(x)</span>=<span>1<span>3−0</span></span><span>(8<span>(<span><span>12</span><span>x2</span><span>]<span>30</span></span></span>)</span>+<span>∫<span>30</span></span>−6dx<span>)</span></span></span>
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F=1.8C+32

We are given C=-20 degrees

plugging this value in the given equation

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Answer:

A. 6

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C. The words "less than", "equal to", and "no more than"

Step-by-step explanation:

For a, we would want to first make an equation to represent the problem. It appears you've already figured that out though, so lets solve it!

Subtract 56 on both sides.

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Now divide -36 by -6 to isolate m.

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