Use the definition of the definite integral to evaluate \( \int_{0}^{4}\left(x^{2}-5\right) d x \). Evaluate the right Riemann Sum. Choose the correct answer below. A. \( \int_{a}^{\text {b }} f(x) d x=\lim _{\Delta \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\left(\frac{n}{4 k}\right)^{2}-5\right) \frac{n}{4} \) B. \( f(x) d x=\lim _{\Delta \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\left(\frac{4 k}{n}\right)^{2}-5\right) \frac{4}{n} \) C. \( \int_{a}^{b} f(x) d x=\lim _{\Delta \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\left(\frac{n}{4}\right)^{2}-5\right) \frac{n}{4 k} \) D. \( \int_{a}^{b} f(x) d x=\lim _{\Delta \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\left(\frac{4}{n}\right)^{2}-5\right) \frac{4 k}{n} \) Fvalıato the interral

To evaluate the definite integral \( \int_{0}^{4}\left(x^{2}-5\right) d x \) using the definition of the definite integral, we will follow these steps: 1. **Identify the function and interval**: The function is \( f(x) = x^2 - 5 \) and the interval is from \( a = 0 \) to \( b = 4 \). 2. **Set up the Riemann sum**: We will use the right Riemann sum to approximate the integral. The interval \([0, 4]\) will be divided into \( n \) subintervals, each of width \( \Delta x = \frac{b - a}{n} = \frac{4 - 0}{n} = \frac{4}{n} \). 3. **Determine the right endpoints**: The right endpoint of the \( k \)-th subinterval is given by: \[ x_k = a + k \Delta x = 0 + k \cdot \frac{4}{n} = \frac{4k}{n} \] 4. **Evaluate the function at the right endpoints**: We need to evaluate \( f(x_k) \): \[ f\left(x_k\right) = f\left(\frac{4k}{n}\right) = \left(\frac{4k}{n}\right)^2 - 5 = \frac{16k^2}{n^2} - 5 \] 5. **Set up the Riemann sum**: The Riemann sum is given by: \[ \sum_{k=1}^{n} f\left(x_k\right) \Delta x = \sum_{k=1}^{n} \left(\frac{16k^2}{n^2} - 5\right) \cdot \frac{4}{n} \] 6. **Combine the terms**: This becomes: \[ \sum_{k=1}^{n} \left(\frac{16k^2}{n^2} - 5\right) \cdot \frac{4}{n} = \sum_{k=1}^{n} \left(\frac{64k^2}{n^3} - \frac{20}{n}\right) \] 7. **Take the limit as \( n \to \infty \)**: The definite integral is then: \[ \int_{0}^{4} (x^2 - 5) \, dx = \lim_{n \to \infty} \sum_{k=1}^{n} \left(\frac{64k^2}{n^3} - \frac{20}{n}\right) \] Now, let's evaluate the limit and find the correct option from the choices provided. ### Evaluating the Integral We can also directly evaluate the integral: \[ \int_{0}^{4} (x^2 - 5) \, dx = \left[ \frac{x^3}{3} - 5x \right]_{0}^{4} \] Calculating this: 1. Evaluate at \( x = 4 \): \[ \frac{4^3}{3} - 5 \cdot 4 = \frac{64}{3} - 20 = \frac{64}{3} - \frac{60}{3} = \frac{4}{3} \] 2. Evaluate at \( x = 0 \): \[ \frac{0^3}{3} - 5 \cdot 0 = 0 \] Thus, the definite integral evaluates to: \[ \int_{0}^{4} (x^2 - 5) \, dx = \frac{4}{3} - 0 = \frac{4}{3} \] ### Choosing the Correct Option Now, let's analyze the options provided: - **Option A**: Incorrect, as it uses \( \frac{n}{4k} \). - **Option B**: Correct, as it uses \( \left(\frac{4k}{n}\right)^2 - 5 \) and \( \Delta x = \frac{4}{n} \). - **Option C**: Incorrect, as it uses \( \frac{n}{4} \). - **Option D**: Incorrect, as it uses \( \left(\frac{4}{n}\right)^2 - 5 \) incorrectly. Thus, the correct answer is **B**: \[ \int_{a}^{b} f(x) d x=\lim _{\Delta \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\left(\frac{4 k}{n}\right)^{2}-5\right) \frac{4}{n} \]