Set up an integral representing the area A of the region enclosed by the given curves. y = 3x, y = 4x, x = 1

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We have been given a region bounded by 3 lines `y=3x` , `y =4x` and x=1. First of all, we will graph the region bounded by 3 curves.

We can see that the first two equations are in slope-intercept form with intercept zero. So these two lines will pass through...

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We have been given a region bounded by 3 lines `y=3x` , `y =4x` and x=1. First of all, we will graph the region bounded by 3 curves.

We can see that the first two equations are in slope-intercept form with intercept zero. So these two lines will pass through the origin. We can see that slope of both the lines are different, which means one line will be above another line.

Next, we will identify which line is the upper line and which line is the lower line.

The line with greater slope will be upper line and line with smaller slope will be lower line. We can see that slope of 4 is greater than 3, so line y = 4x will be above line y = 3x. I am sharing the bounded region with the attached image.

Since our region is bounded between upper curve and lower curve, we will use integral a to b (Upper curve - Lower Curve)dx.

`int_a^b(upper curve- Lower Curve )dx`

Our next step is to identify the bounds for our integral.

We can see that the region is bounded in 1st quadrant, so it starts from `x=0`. So these will be bounds for our integral.

Please go ahead and substitute these bounds and equations of upper and lower curve in given set-up shared above before looking at the next step.` `

`int_0^1(4x)-(3x)dx`

Therefore, the above integral represents area A bounded by the given curves.

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