evaluate the integral.\n int_{0}^{1} (19x^{e}+17e^{x}) dx

Explanation:

Step1: Use integral - sum rule

$\int_{0}^{1}(19x^{e}+17e^{x})dx=\int_{0}^{1}19x^{e}dx+\int_{0}^{1}17e^{x}dx$

Step2: Apply power - rule for $\int_{0}^{1}19x^{e}dx$

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). So, $\int_{0}^{1}19x^{e}dx=19\int_{0}^{1}x^{e}dx=19[\frac{x^{e + 1}}{e + 1}]_{0}^{1}=\frac{19}{e + 1}(1^{e+1}-0^{e + 1})=\frac{19}{e + 1}$

Step3: Apply exponential - rule for $\int_{0}^{1}17e^{x}dx$

The integral of $e^{x}$ is $e^{x}+C$. So, $\int_{0}^{1}17e^{x}dx=17[e^{x}]_{0}^{1}=17(e^{1}-e^{0})=17(e - 1)$

Step4: Combine the results

$\int_{0}^{1}(19x^{e}+17e^{x})dx=\frac{19}{e + 1}+17(e - 1)=\frac{19+17(e - 1)(e + 1)}{e + 1}=\frac{19+17(e^{2}-1)}{e + 1}=\frac{19+17e^{2}-17}{e + 1}=\frac{17e^{2}+2}{e + 1}$

Answer:

$\frac{17e^{2}+2}{e + 1}$