## Calculus: Early Transcendentals (2nd Edition)

Using Riemann sums, the definition of the definite integral says that this integral is equal to the limit when the width of the arbitrary partition $\Delta$ goes to $0$ of the sum $\sum_{i=0}^Nf(x_i^*)\Delta x_i$ where $x_i^*\in(x_{i-1},x_i)$, and $x_i$ are the points defining the partition. Since $\Delta x_i$ are positive and the function $f$ is negative everywhere,which means that every $f(x_i^*)<0$, then every member of the sum $f(x_i^*)\Delta x_i<0$ so the total sum must be negative, and so is the integral. Another way of clearly saying this, is that there is some $M<0$ such that $f(x)\leq M$ for every $x$ so $$\sum_{i=0}^Nf(x_i^*)\Delta x_i\leq\sum_{i=0}^NM\Delta x_i=M\sum_{i=0}^N\Delta x_i=M(b-a)<0,$$ because $M<0$ and $b>a$.