Question : Representing the area by integral, find the area of the region bounded by the lines $\displaystyle x=4, y=1$ and $\displaystyle x+4y = 4$.
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Solution:
Question : Representing the area by integral, find the area of the region bounded by the lines $\displaystyle x=4, y=1$ and $\displaystyle x+4y = 4$.
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Solution:
I just notice that you said "representing the area by an integral". wau wau's answer is best. But if it were not for that requirement, that's a right triangle: area= (1/2)height * base. Find the points where x+ 4y= 4 and you are almost done. You can use that to check your work.
Find the area under the top function and subtract by the area under the bottom function.
$\displaystyle \int_{0}^{4}1dx-\left |\int_{0}^{4}\frac{x-4}{4}dx \right |=2$
Any vertical line restraint only represents were to stop.
x=4 then
$\displaystyle \int_{n}^{4}$
---$\displaystyle \int$
I told you earlier, in post number 3 how you could easily check this yourself. This is a right triangle. It's area is just (1/2)height times width: one leg is from y= 0 to y= 1 so has height 1, the other is from x= 0 to x= 4 so has width 4. The area is (1/2)(1)(4)= 2.