Question

let $f$ be a continuous function such that $\int_{7}^{9} f(x) \\, dx = -4$ and $\int_{6}^{9} f(x) \\, dx = -14$. what is the value of $\int_{6}^{7} \frac{1}{2}f(x) \\, dx$?

Explanation:

Step1: Use the property of definite integrals (additivity)

We know that \(\int_{6}^{9}f(x)dx=\int_{6}^{7}f(x)dx+\int_{7}^{9}f(x)dx\). We can solve for \(\int_{6}^{7}f(x)dx\) by rearranging this formula.
Given \(\int_{7}^{9}f(x)dx = - 4\) and \(\int_{6}^{9}f(x)dx=-14\), then \(\int_{6}^{7}f(x)dx=\int_{6}^{9}f(x)dx-\int_{7}^{9}f(x)dx\)
Substitute the known values: \(\int_{6}^{7}f(x)dx=-14 - (-4)=-14 + 4=-10\)

Step2: Use the constant multiple rule for definite integrals

The constant multiple rule states that \(\int_{a}^{b}kf(x)dx = k\int_{a}^{b}f(x)dx\) for a constant \(k\).
We want to find \(\int_{6}^{7}\frac{1}{2}f(x)dx\), using the constant multiple rule with \(k = \frac{1}{2}\) and \(\int_{6}^{7}f(x)dx=-10\), we get:
\(\int_{6}^{7}\frac{1}{2}f(x)dx=\frac{1}{2}\int_{6}^{7}f(x)dx\)
Substitute \(\int_{6}^{7}f(x)dx=-10\) into the formula: \(\frac{1}{2}\times(-10)=-5\)

Answer:

\(-5\)