# Let g(x)=90/(3x+4), for 2≤x≤12. Find the area enclosed by the lines L,x=2,x=12 and the graph of h.The graph of g is reflected in the x-axis to give the graph of h.The area of the region...

Let g(x)=90/(3x+4), for 2≤x≤12. Find the area enclosed by the lines L,x=2,x=12 and the graph of h.

The graph of g is reflected in the x-axis to give the graph of h.The area of the region enclosed by the lines L,x=2,x=12 and the x-axis is 120cm^2

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Let g(x)=90/(3x+4), for 2≤x≤12. Find the area enclosed by the lines L,x=2,x=12 and the graph of h.

The graph of g is reflected in the x-axis to give the graph of h.The area of the region enclosed by the lines L,x=2,x=12 and the x-axis is 120cm^2

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Ok, start with a graph!

Here's g and its reflection (h). Let's find the area between h and the x-axis, and just add that to 120. It doesn't matter what L looks like (it could be a horizontal line at y = 12 for all we care), since we already know the area between L and the x-axis.

So you need to find `int abs(h) dx = int g(x) dx` from 2 to 12.

`int 90/(3x+4) dx = 30 int 3/(3x+4) = 30 ln(3x + 4) `

Now evaluate at 12 and 2, and find the difference: 41.589

So that's the area below the x-axis (usually this value is negative, but we're dealing with square centimeters, so use the absolute value). Add on the area above the x-axis and you have

161.589 cm^2