If for $a\le x\le b$, $\frac{d}{dx}f(x)=g(x)$ then $\int_{a}^{b}f(x)g(x)dx=$

Question

Accepted Answer

To address the question, we need to evaluate the integral $\int_{a}^{b} f(x)g(x) \, dx$ given that $\frac{d}{dx}f(x) = g(x)$. The correct answer is given as Option 4: $\frac{{\{f(b)\}^{2}-{\{f(a)\}}^{2}}{2}$.

### Explanation for Option 4:

Option 4 states that the integral $\int_{a}^{b} f(x)g(x) \, dx$ equals $\frac{{\{f(b)\}^{2}-{\{f(a)\}}^{2}}{2}$. This can be understood by considering the relationship between the functions $f(x)$ and $g(x)$. Since $\frac{d}{dx}f(x) = g(x)$, $g(x)$ is the derivative of $f(x)$.

The expression $\frac{{\{f(b)\}^{2}-{\{f(a)\}}^{2}}{2}$ can be derived from the integration by parts formula or by recognizing it as a specific result of integrating the product of a function and its derivative over an interval. This result is consistent with the integral of a function squared, which is a common result in calculus when dealing with such expressions.

### Explanation for why other options are incorrect:

1. **Option 1: $g(b) - g(a)$**

This option suggests that the integral of the product of a function and its derivative over an interval is simply the difference in the values of the derivative at the endpoints. This is incorrect because it does not account for the integration process, which involves accumulating the area under the curve of the product $f(x)g(x)$. The expression $g(b) - g(a)$ would be more relevant if we were dealing with the integral of $g(x)$ alone, not the product $f(x)g(x)$.

2. **Option 2: $f(b) - f(a)$**

This option implies that the integral of the product $f(x)g(x)$ is equivalent to the difference in the values of $f(x)$ at the endpoints. This is incorrect because $f(b) - f(a)$ represents the net change in $f(x)$ over the interval $[a, b]$, which is not directly related to the integral of the product of $f(x)$ and its derivative.

3. **Option 3: $\frac{{\{g(b)\}^{2}-{\{g(a)\}}^{2}}{2}$**

This option suggests that the integral is related to the square of the derivative values at the endpoints. However, this does not align with the integration of the product $f(x)g(x)$. The expression $\frac{{\{g(b)\}^{2}-{\{g(a)\}}^{2}}{2}$ would be more relevant in a context where the integral directly involves the squares of the derivative values, which is not the case here.

In conclusion, Option 4 is the correct answer because it accurately represents the integral of the product of a function and its derivative over an interval, consistent with known results in calculus.