Question

1. suppose f has absolute minimum value m and absolute maximum value m. between what two values must (int_{0}^{2}f(x)dx) lie? which property of integrals allows you to make your conclusion?

Explanation:

Step1: Recall integral property

If \(m\leq f(x)\leq M\) for \(a\leq x\leq b\), then \(m(b - a)\leq\int_{a}^{b}f(x)dx\leq M(b - a)\). Here \(a = 0\), \(b = 2\).

Step2: Apply the property

Substitute \(a = 0\), \(b = 2\) into the inequality. We get \(m(2-0)\leq\int_{0}^{2}f(x)dx\leq M(2 - 0)\), which simplifies to \(2m\leq\int_{0}^{2}f(x)dx\leq2M\).

Answer:

The value of \(\int_{0}^{2}f(x)dx\) must lie between \(2m\) and \(2M\), using the property that if \(m\leq f(x)\leq M\) on the interval \([a,b]\), then \(m(b - a)\leq\int_{a}^{b}f(x)dx\leq M(b - a)\).