Evaluate the following definite integral: $\int_{-3}^3(3x + 5)\,dx$ I'm new to the topic of integration and I'm struggling big time with it.
I am trying to evaluate $\int_{-3}^3(3x + 5)\,dx$.
Any help is greatly appreciated.
Thank you

If you've learned the power rule for integration, then this will serve to refresh that learning. If you haven't learned the power rule, then you'll find this very helpful.
In general, if we need to integrate a polynomial, we use the rule:
$$\int_a^b {cx^n} \,dx = c\cdot\dfrac{x^{n+1}}{n+1}\Big|_a^b, \quad \forall n\in \mathbb Z, n\neq -1$$
In your case, we have the definite integral $$\int_{-3}^3 3x + 5 \,dx = \int_{-3}^3 3x^1 + 5x^0 \,dx = \int_{-3}^3 3x^1 \,dx + \int_{-3}^3 5x^0\,dx$$
Using the power rule on the right-hand side, and then evaluating the result at (upper-bound $-$ lower-bound) gives us $$\begin{align} \int_{-3}^3 3x^1 \,dx + \int_{-3}^3 5x^0\,dx & = 3\dfrac{x^{1+1}}{(1+1)}\Big|_{-3}^3 + 5\cdot \dfrac{x^{(0+1)}}{(0+1)}\Big|_{-3}^3 \\ \\ &= 3\cdot\dfrac{x^2}{2}\Big|_{-3}^3 + 5x\Big|_{-3}^3 \\ \\ & = \dfrac 32 ((3)^2 - (-3)^2) + 5(3 - (-3)) \\ \\ & = 0 + 5\cdot 6 \\ \\ &= 30\end{align}$$
