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R is the first quadrant region enclosed by the x-axis, the curve y=2x+a, and the line x=a, where a>0. Find the value of a so that the area of the region R is 18 square units. A.3 B.3.772 C. 4.242 D. 9
The x-intercept of y=2x+a is x = -a/2 So, you have a triangle with base 3a/2 and height 3a, so the area is 9a^2/4. If \[ 9a^2/4 = 18, a = √8 \] check: \[∫[-a/2,a] 2x+a dx = 9a^2/4\]
sqrt 8 isnt an answer choice @Ineedhelplz
...hmm
let's see if he can help
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@mrm
Hi guys. Let me see what you have done so far.
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