7. Let A(t) be the area of the region in the first quadrant enclosed by the coordinate axes, the curve y=e^(-x), and the vertical line x=t, t\u003e0. Let V(t) be the volume of the solid generated by revolving the region about the x-axis. Find the following limits.a. lim(x→∞)A(t)

Ch. 7 - Transcendental Functions

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Ch. 7 - Transcendental Functions

Problema 7.AAE.7a

Capitolo 7, Problema 7.AAE.7a

7. Let A(t) be the area of the region in the first quadrant enclosed by the coordinate axes, the curve y=e^(-x), and the vertical line x=t, t>0. Let V(t) be the volume of the solid generated by revolving the region about the x-axis. Find the following limits.
a. lim(x→∞)A(t)

Guida verificata passo dopo passo

First, understand the region described: it is bounded by the x-axis (y=0), the y-axis (x=0), the curve \(y = e^{-x}\), and the vertical line \(x = t\) where \(t > 0\).

To find the area \(A(t)\) of this region, set up the definite integral of the function \(y = e^{-x}\) from \(x=0\) to \(x=t\): \[A(t) = \int_0^t e^{-x} \, dx\]

Evaluate the integral symbolically (do not compute the final value yet): \[\int e^{-x} \, dx = -e^{-x} + C\] So, \[A(t) = [-e^{-x}]_0^t = (-e^{-t}) - (-e^{0}) = 1 - e^{-t}\]

Now, to find the limit as \(t\) approaches infinity, consider the behavior of \(e^{-t}\) as \(t \to \infty\). Since \(e^{-t}\) approaches 0, the expression \(1 - e^{-t}\) approaches 1.

Therefore, the limit is \[\lim_{t \to \infty} A(t) = 1\] This represents the total area under the curve \(y = e^{-x}\) from 0 to infinity in the first quadrant.

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