Evaluate
$\int (x^3 + x^5 + x^7 + x^9 + x^{11}) \cdot e^{x^{13}} \cdot \sin(4x^3) \cdot \cos(6x^5) , dx$.
I am not sure exactly where to start here; i tried evaluating using numerical quadrature, applying the trapezoidal rule, as such:
I chose the range of integration by determining the lower and upper bounds of integration, denoted as $a$ and $b$, respectively. I then divided the interval $[a, b]$ into $N$ subintervals of equal width. Let $\Delta x$ represent the width of each subinterval, given by $\Delta x = \frac{{b - a}}{{N}}$. After, I calculated the value of the integrand at each node within the subintervals. In this case, the nodes are the endpoints of the subintervals and can be represented as $x_i = a + i \Delta x$, where $i$ ranges from 0 to $N$.
Next, I applied the trapezoidal rule to approximate the integral by summing the weighted values of the function at the nodes. The trapezoidal rule states that the integral within each subinterval can be approximated as the average of the function values at the endpoints, multiplied by the width of the subinterval.
Approximation within each subinterval: $$ \int_{x_i}^{x_{i+1}} (x^3 + x^5 + x^7 + x^9 + x^{11}) \cdot e^{x^{13}} \cdot \sin(4x^3) \cdot \cos(6x^5) \, dx \approx \frac{{\Delta x}}{{2}} \left[ f(x_i) + f(x_{i+1}) \right], $$ where $f(x_i)$ represents the value of the integrand at $x_i$.
Finally I computed the sum of the weighted values obtained from the previous step across all subintervals to obtain the numerical approximation of the integral.
Numerical approximation: $$ \int_{a}^{b} (x^3 + x^5 + x^7 + x^9 + x^{11}) \cdot e^{x^{13}} \cdot \sin(4x^3) \cdot \cos(6x^5) \, dx \approx \sum_{i=0}^{N-1} \frac{{\Delta x}}{{2}} \left[ f(x_i) + f(x_{i+1}) \right].$$
Did I do something wrong? Is this the correct approximation?