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
 A: Have you learned the Fundamental Theorem of Calculus yet?
If so, recall that you can evaluate definite integrals of $3x + 5$ using an antiderivative. In particular
$$
\int_{-3}^3 3x + 5 \,dx = \left[\frac{3}{2}x^2 + 5x \right]_{-3}^3
$$
Otherwise, you should draw a graph of this function and use simple geometry to find the integral - recall that the integral is related to the area between the function and the $x$-axis; if the area is above the $x$-axis, you count it positively, and if it is below, you count it negatively.
Further, because this function is so basic, you'll be able to calculate the areas involved using simple geometric formulae for triangles.

A: 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 $$\begin{align} \int_{-3}^3 3x^1 \,dx + \int_{-3}^3 5x^0\,dx & = 3\dfrac{x^{1+1}}{(1+1)}\Big|_{-3}^3 + 5\cdot \dfrac{x^{(0+1)}}{(0+1)}\Big|_{-3}^3 \\ \\ &= 3\cdot\dfrac{x^2}{2}\Big|_{-3}^3 + 5x\Big|_{-3}^3 \\ \\ & = \dfrac 32 ((3)^2 - (-3)^2) + 5(3 - (-3)) \\ \\ & = 0 + 5\cdot 6 \\ \\ &= 30\end{align}$$
A: so we have
$\int{(3 \cdot x+5)}dx=3 \cdot x^2/2+5 \cdot x$
now  we insert bounds  $3$ and $-3$, we will do separately  when we insert bound values  and denote them by $A_1$ and $A_2$
$A_1=3 \cdot (3^2)/2+5 \cdot 3=3\cdot9/2+15$
$A_2=3 \cdot (-3)^2/2-15$
now total result is $A_1-A_2$
A: Integration of a Variable
Integration requires you to add $1$ to the power (exponent value) of the variable and then divide by that sum. (LHS) Below the power of the variable, $x$, is $1$, so you would add $1+1=2$ to get $x$ raised to the power of $2$ and then divide the entire term by $2$. (RHS)
$$ \int n^x = \frac{n^{x+1}}{x+1} \implies \int x \space dx= \frac{x^2}{2} + c$$
Integration of a Constant
With constants what you really have is $ax^0$ which is simply the constant itself, $a$, (anything raised to the power of $0$ is $1$). So following from above, when you add $1$, you get $1+0=1$ to get $x$ raised to the power of $1$ which is just $x$. So therefore the integral of any constant is the constant itself times $x$. See below.
$$\int a \space dx= ax + c$$
Applying those rules to the question:
$$\int_{-3}^3 3x + 5 \space dx = \left[\frac{3}{2} x^2 + 5x \right]_{-3}^3 = \left(\frac{3}{2} (3)^2 + 5(3) \right) - \left(\frac{3}{2} (-3)^2 + 5(-3)  \right) = \cdots$$
With definite integrals once you've calculated the integral, you substitute the lower and upper limit into the integral and calculate (subtracting the lower limit from the upper limit). From then on its just simple arithmetic. 
