Question

If $f$ and $g$ are integrable on $[a, b]$ and $f(x) \geq g(x)$ for all $x$ in $[a, b]$, then

• A. $\int_a^b f(x) \, dx \neq \int_a^b g(x) \, dx$
• B. $\int_a^b f(x) \, dx \geq \int_a^b g(x) \, dx$
• C. $\int_a^b f(x) \, dx \leq \int_a^b g(x) \, dx$
• D. $\int_a^b f(x) \, dx$ and $\int_a^b g(x) \, dx$ does not exist

Hint:

Integration is a monotonic operator. If one curve lies completely above another curve over an interval, the area under the higher curve must be greater than or equal to the area under the lower curve.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The comparison property of definite integrals states that integration preserves inequalities between functions.

Step 2: Detailed Explanation:
Let $f(x)$ and $g(x)$ be two integrable functions on the interval $[a, b]$ such that: \[ f(x) \geq g(x) \text{ for all } x \in [a, b] \]
This inequality implies that: \[ f(x) - g(x) \geq 0 \text{ for all } x \in [a, b] \]
Since the function $h(x) = f(x) - g(x)$ is non-negative on $[a, b]$, the area under the curve $h(x)$ must also be non-negative: \[ \int_a^b h(x) \, dx \geq 0 \implies \int_a^b (f(x) - g(x)) \, dx \geq 0 \]
Using the linearity property of integration, we can split this into two separate integrals: \[ \int_a^b f(x) \, dx - \int_a^b g(x) \, dx \geq 0 \] \[ \int_a^b f(x) \, dx \geq \int_a^b g(x) \, dx \]
Thus, the integral of $f(x)$ is greater than or equal to the integral of $g(x)$.

Step 3: Final Answer:
The relationship is $\int_a^b f(x) \, dx \geq \int_a^b g(x) \, dx$.