Basic integral calculus problems

Short answer: An anti-derivative. That is, if $$\int f(x) \, dx = F(x)$$ then $$F'(x) = f(x).$$

You check if $F'(x) = f(x).$

$$\int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx.$$

It is that $$\int u^n \, du = \dfrac{u^{n+1}}{n + 1} + C,$$ where $C$ is an arbitrary constant.

$$\int_a^b f(x) \, dx = F(b) - F(a).$$

Using the power rule, we get $$ \int x^3 \, dx = \dfrac{1}{4} x^4+ C,$$ where $C$ is an arbitrary constant.

\begin{eqnarray} \int 5 x^4 \, dx &=& 5 \int x^4\, dx \\ \ &=& 5 \dfrac{x^5}{5} + C \\ \ &=& x^5 + C, \end{eqnarray} where $C$ is an arbitrary constant.

\begin{eqnarray} \int_{-1}^3 17\, dx &=& \big[17 x\big]_{-1}^3 \\ \ &=& 17 \left( 3 - \left(-1\right) \right) \\ \ &=& 17 (4) \\ \ &=& 68. \end{eqnarray}

\begin{eqnarray} \int_{-2}^2 3 u^7\, du &=& 3 \left[\dfrac{u^8}{8}\right]_{-2}^2 \\ \ &=& \dfrac{3}{8} \left(2^8 - (-2)^8 \right) \\ \ &=& 0. \end{eqnarray}

\begin{eqnarray} \int (x^3 - 5x) \, dx &=& \int x^3 \, dx - 5 \int x \, dx \\ \ &=& \dfrac{1}{4}x^4 -\dfrac{5}{2} x^2 + C, \end{eqnarray} where $C$ is an arbitrary constant.

From the previous question, we know that $$\int (x^3 - 5x) \, dx = \dfrac{1}{4}x^4 -\dfrac{5}{2} x^2 + C,$$ where $C$ is an arbitrary constant. $$ $$ So, \begin{eqnarray} \int_1^3 (x^3 - 5x) \, dx &=& \left[ \dfrac{1}{4}x^4 -\dfrac{5}{2} x^2 \right]_1^3 \\ \ &=& \left[ \dfrac{x^2}{4} \left( x^2 - 10 \right) \right]_1^3 \\ \ &=& \dfrac{3^2}{4} \cdot \left( 3^2 - 10 \right) - \dfrac{1^2}{4} \cdot \left( 1^2 - 10 \right) \\ \ &=& \dfrac{9 \cdot (-1)}{4} - \dfrac{1 \cdot (-9)}{4} \\ \ &=& 0. \end{eqnarray}

$$\int y^{3/2} \, dy = \dfrac{2}{5}y^{5/2} + C$$

\begin{eqnarray} \int_0^4 3 \sqrt{x}\, dx &=& 3 \int_0^4 x^{1/2} \, dx \\ \ &=& \dfrac{3}{3/2} \left[x^{3/2}\right]_0^4 \\ \ &=& 2 \left(4^{3/2} - 0^{3/2} \right) \\ \ &=& 16. \end{eqnarray}