Question

The area under the graph of f over the interval [-2,2]
F(x)=( 5-x^2, if x<0 if x less than or equal to 0
( 5,

Answer 1

To find the area under the graph of \(f(x)\) over the interval \([-2,2]\), we need to evaluate the definite integral of \(f(x)\) from \(-2\) to \(2\).

Since \(f(x)\) is defined piece-wise, we have two different expressions for \(f(x)\) based on the value of \(x\):

For \(x<0\), \(f(x) = 5 - x^2\).
For \(x \geq 0\), \(f(x) = 5\).

We can split the interval \([-2,2]\) into two parts: \([-2,0]\) and \([0,2]\). In each part, we use the corresponding expression for \(f(x)\).

For \([-2,0]\):
\[
\int_{-2}^{0}(5 - x^2)dx
\]

We can integrate \(5 - x^2\) with respect to \(x\) by following these steps:
1. Distribute the integral sign:
\[
\int_{-2}^{0}5dx - \int_{-2}^{0}x^2dx
\]

2. Integrate \(5\) with respect to \(x\). Since \(5\) is a constant, it becomes \(5x\) when integrated:
\[
5x
\]

3. Integrate \(x^2\). The antiderivative of \(x^2\) is \(\frac{x^3}{3}\):
\[
\frac{x^3}{3}
\]

4. Evaluate the expression at the bounds of integration:
\[
\left[\frac{x^3}{3}\right]_{-2}^{0} = \frac{0^3}{3} - \frac{(-2)^3}{3} = 0 - (-\frac{8}{3}) = \frac{8}{3}
\]

So, the area under the graph of \(f(x)\) over the interval \([-2,0]\) is \(\frac{8}{3}\).

For \([0,2]\):
\[
\int_{0}^{2}5dx
\]

We can integrate \(5\) with respect to \(x\) in the same way as in the previous step:
\[
\int_{0}^{2}5dx = 5x\bigg|_{0}^{2} = 5(2) - 5(0) = 10
\]

So, the area under the graph of \(f(x)\) over the interval \([0,2]\) is \(10\).

Therefore, the total area under the graph of \(f(x)\) over the interval \([-2,2]\) is the sum of the areas of the two intervals:
\[
\frac{8}{3} + 10 = \frac{34}{3}
\]

In conclusion, the area under the graph of \(f(x)\) over the interval \([-2,2]\) is \(\frac{34}{3}\).