Let R be the region in the first quadrant that is enclosed by the graph of y=tan x, the x-axis, and the line x = pi/3
a) Find the area of R
b) Find the volume of the solid formed by revolving R about the x-axis
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a) Integrate tanx dx from 0 to pi/3
b) Integrate (pi) tan^2 x dx from 0 to pi/3.
a) Draw the photograph. First, your area is limited from the Y axis to an x of two. between the vertical Y axis and the vertical line by an x of two. it is from x = 0 to x = 2. So permit 0 and a pair of be the endpoints of your specific quintessential. combine y = 2e^x + 3x on that era. so which you have the quintessential from 0 to 2 of (2e^x + 3x) dx. Or, 2 circumstances the quintessential of e^x plus thrice the quintessential of x. Or: 2(e^x) plus 3[(x^2)/2]. evaluate it first with x = 2. that must be 2(e^2) plus 3[(2^2)/2]. Or 2e^2 + 6. Then evaluate it with x = 0, and subtract to get the adaptation. evaluate it with x = 0 for this reason: 2e^0 + 3/2 (0)^2 = 2(a million) + 0. So the evaluated quintessential is 2e^2 + 6 - 2 or 2e^2 + 4. it is the section enclosed between the curve and the X axis, and between the Y axis and the vertical line at x = 2. section = (2e^2 + 4) sq. instruments. To get the quantity you utilize shells or jewelry or some such subject. it truly is a reachable technique--is smart and boils all the way down to basic formula. lower back, attempt to entice the photograph.
Ah, the joys of integral calculus - do it your self!