Volume of Perpendicular Slices for Region Bounded by y=e^x and x=1

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finding volume by integration

Homework Statement



Consider the region bounded by y=e^x, the x-axis, the y-axis, and the line x = 1. A solid is created so that the given region is its base and cross-sections perpendicular to the x-axis are squares. What is the volume of a slice perpendicular to the x-axis?


The Attempt at a Solution



since the regions is bounde by y=e^2, y=0. and x=1,
doesn't the volume of a slice perpendicular to the x-axis
has to be (e^x)(e^x)dx ..(since the base and height are equal)

so the answer should be [Integral from 0 to 1](e^(2x)) dx


Am I wrong? I thought I set it up right but the webassign thing's
keel saying that somethings wrong

I'm in a hurry so please help!
amd
 
Last edited:
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That looks correct to me.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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