\( f(x)=x^{3}+2 x^{2}+x+1 \) In tervalo \( [0,5] \) \( =50 \)

It seems like you're asking for the evaluation of the function \( f(x) = x^3 + 2x^2 + x + 1 \) over the interval \([0, 5]\). However, it's not clear what you mean by "=50". If you want to find the definite integral of \( f(x) \) from 0 to 5, we can calculate that. The definite integral will give us the area under the curve of the function from \( x = 0 \) to \( x = 5 \). To find the definite integral, we first need to find the antiderivative of \( f(x) \): \[ \int f(x) \, dx = \int (x^3 + 2x^2 + x + 1) \, dx \] Calculating the antiderivative: \[ \int f(x) \, dx = \frac{x^4}{4} + \frac{2x^3}{3} + \frac{x^2}{2} + x + C \] Now, we evaluate this from 0 to 5: \[ \int_0^5 f(x) \, dx = \left[ \frac{x^4}{4} + \frac{2x^3}{3} + \frac{x^2}{2} + x \right]_0^5 \] Calculating at \( x = 5 \): \[ = \frac{5^4}{4} + \frac{2 \cdot 5^3}{3} + \frac{5^2}{2} + 5 \] Calculating each term: 1. \( \frac{5^4}{4} = \frac{625}{4} = 156.25 \) 2. \( \frac{2 \cdot 5^3}{3} = \frac{2 \cdot 125}{3} = \frac{250}{3} \approx 83.33 \) 3. \( \frac{5^2}{2} = \frac{25}{2} = 12.5 \) 4. \( 5 = 5 \) Now, summing these values: \[ \int_0^5 f(x) \, dx = 156.25 + 83.33 + 12.5 + 5 \] Calculating the total: \[ = 156.25 + 83.33 = 239.58 \] \[ 239.58 + 12.5 = 252.08 \] \[ 252.08 + 5 = 257.08 \] Thus, the definite integral of \( f(x) \) from 0 to 5 is approximately \( 257.08 \). If you meant something else by "=50", please clarify!