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Find the area of the region between the x-axis and the curve(a) y = e−3x for x ≥ 0. (b) y = 8x2−4 for x ≥ 4

Question

Find the area of the region between the x-axis and the curve

(a) y=e3x y = e^{-3x} for x0 x \geq 0 .
(b) y=8x24 y = 8x^2 - 4 for x4 x \geq 4

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Solution

To find the area of the region between the x-axis and the curve, we need to calculate the definite integral of the given functions.

(a) For the function y = e^(-3x) for x ≥ 0, we need to find the integral of this function from x = 0 to infinity. However, since the function approaches zero as x approaches infinity, we can find the integral from x = 0 to a finite value and then take the limit as that value approaches infinity.

Let's find the integral of y = e^(-3x) from x = 0 to a finite value, say x = b: ∫[0 to b] e^(-3x) dx

To solve this integral, we can use the power rule for integration. The integral of e^(-3x) is (-1/3)e^(-3x). So, the integral becomes: (-1/3) ∫[0 to b] e^(-3x) dx = (-1/3) * [e^(-3x)] evaluated from 0 to b

Now, let's evaluate this expression at the limits: (-1/3) * [e^(-3b) - e^0]

As b approaches infinity, e^(-3b) approaches zero, and e^0 is equal to 1. Therefore, the expression simplifies to: (-1/3) * (0 - 1) = 1/3

So, the area between the x-axis and the curve y = e^(-3x) for x ≥ 0 is 1/3.

(b) For the function y = 8x^2 - 4 for x ≥ 4, we need to find the integral of this function from x = 4 to infinity. However, since the function approaches infinity as x approaches infinity, we cannot find a finite value for the integral.

Therefore, the area between the x-axis and the curve y = 8x^2 - 4 for x ≥ 4 is infinite.

This problem has been solved

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