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Consider the region bounded by the graph of $y = 3x + 2$, the $x$-axis, and the lines $x = 5$, $x = 8$. A reasonable upper bound for the area of this region is

          Consider the region bounded by the graph of $y = 3x + 2$, the $x$-axis, and the lines $x = 5$, $x = 8$.
A reasonable upper bound for the area of this region is
        
Consider the region bounded by the graph of y = 3x + 2, the x-axis, and the lines x = 5, x = 8.
A reasonable upper bound for the area of this region is

Added by Magdalena Z.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Consider the region bounded by the graph of y=3x+2, the x-axis, and the lines x=5,x=8. A reasonable upper bound for the area of this region is Consider the region bounded by the graph of y3+2,the -axis,and the lines =5,=8 A reasonable upper bound for the area of this region is
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Transcript

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00:03 We want to graph these functions and find the area between their curves.
00:10 So here's the graph of the two functions.
00:13 We're looking for the area of this region in here.
00:16 We're asked to approximately find the area.
00:19 So we can see that the blue linear function is greater than this x to the fourth function from x is negative 1 .493 all the way up to x's positive 1 .75.
00:35 So we're going to integrate, we're going to make a definite integral, integrate with respect to the x -axis between these two x values, negative 1 .493 and 1 .75.
00:48 So negative 1 .493, negative 1 .493.
01:13 And that's going to go up to x's 1 .785.
01:18 Now, when you find the area between two curves, you take the greater function and subtract the lesser function.
01:34 So the linear minus the x to the fourth function.
01:40 So the linear function, x plus 2, subtract the lesser function, x to the fourth, minus 2x squared.
02:08 All right, so our area is going to equal the integral from negative 1 .493 to 1 .785.
02:29 And let's clean this up a little bit.
02:31 Let's try to write these terms in descending order...
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