Computing areas On the interval [0,2], the graphs of f(x)=x²/3 and g(x)=x²(9−x²)^(-1/2) have similar shapes.
a. Find the area of the region bounded by the graph of f and the x-axis on the interval [0,2].

Verified step by step guidance

1. Identify the function and the interval for which you need to find the area. Here, the function is \(f(x) = \frac{x^2}{3}\) and the interval is \([0, 2]\).
2. Recall that the area under the curve of a function \(f(x)\) from \(a\) to \(b\) is given by the definite integral \(\int_a^b f(x) \, dx\). In this case, you want to compute \(\int_0^2 \frac{x^2}{3} \, dx\).
3. Set up the integral explicitly: \(\int_0^2 \frac{x^2}{3} \, dx = \frac{1}{3} \int_0^2 x^2 \, dx\). You can factor out the constant \(\frac{1}{3}\) from the integral.
4. Find the antiderivative of \(x^2\), which is \(\frac{x^3}{3}\). So, the integral becomes \(\frac{1}{3} \left[ \frac{x^3}{3} \right]_0^2\).
5. Evaluate the definite integral by substituting the upper and lower limits: calculate \(\frac{1}{3} \left( \frac{2^3}{3} - \frac{0^3}{3} \right)\) to express the area bounded by the graph of \(f\) and the x-axis on \([0,2]\).

Key Concepts

Definite Integral as Area Under a Curve

The definite integral of a function over an interval represents the net area between the graph of the function and the x-axis. For a non-negative function, this integral gives the exact area bounded by the curve and the x-axis within the specified limits.

Integration of Polynomial Functions

Integrating polynomial functions involves applying the power rule, which states that the integral of x^n is (x^(n+1))/(n+1) plus a constant. This rule simplifies finding antiderivatives for functions like f(x) = x²/3.

Evaluating Definite Integrals Using Limits

To compute the definite integral, first find the antiderivative, then evaluate it at the upper and lower bounds of the interval. Subtracting these values yields the exact area under the curve between those points.