Find the average value of
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a. y = √3x over [0, 3]

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Identify the function and the interval over which to find the average value. Here, the function is \(y = \sqrt{3x}\) and the interval is \([0, 3]\).

Recall the formula for the average value of a function \(f(x)\) over the interval \([a, b]\): \[\text{Average value} = \frac{1}{b - a} \int_a^b f(x) \, dx\]

Substitute the given function and interval into the formula: \[\text{Average value} = \frac{1}{3 - 0} \int_0^3 \sqrt{3x} \, dx = \frac{1}{3} \int_0^3 (3x)^{1/2} \, dx\]

Simplify the integrand if possible and set up the integral for evaluation: \[\int_0^3 (3x)^{1/2} \, dx = \int_0^3 3^{1/2} x^{1/2} \, dx = \sqrt{3} \int_0^3 x^{1/2} \, dx\]

Evaluate the integral \(\int_0^3 x^{1/2} \, dx\) using the power rule for integration: \[\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C\] Apply this with \(n = \frac{1}{2}\) and then multiply by \(\sqrt{3}\) and finally multiply by \(\frac{1}{3}\) to find the average value.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Average Value of a Function

The average value of a function f(x) over an interval [a, b] is given by (1/(b - a)) times the definite integral of f(x) from a to b. It represents the mean height of the function on that interval.

Definite Integral

A definite integral calculates the net area under the curve of a function between two points a and b. It is essential for finding accumulated quantities and is used here to find the total 'sum' of function values over the interval.

Integration of Power Functions

Integrating functions like y = √(3x) involves rewriting the function in exponent form and applying the power rule for integration, which states ∫x^n dx = (x^(n+1))/(n+1) + C, for n ≠ -1.

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