Answer: See below
Step-by-step explanation:
A.
Let's split the integral into two parts, by the Sum Rule.
[tex]\int\limits {x-4x^3} \, dx[/tex] [split into 2 integrals]
[tex]\int\limits {x} \, dx -\int\limits {4x^3} \, dx[/tex] [solve integral for each part]
[tex]\frac{1}{2} x^2-x^4+C[/tex] [Remember, we need to add C for constant]
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B.
[tex]\int\limits {\frac{1+x}{\sqrt{x} } } \, dx[/tex] [expand into 2 integrals]
[tex]\int\limits {\frac{1}{\sqrt{x} } } \, dx +\int\limits {\frac{x}{\sqrt{x} } } \, dx[/tex] [simplify second integral]
[tex]\int\limits {\frac{1}{\sqrt{x} } } \, dx +\int\limits {\sqrt{x} } \, dx[/tex] [solve integral for each part]
[tex]2\sqrt{x} +\frac{2}{3}x^3^/^2+C[/tex]
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C.
[tex]\int\limits^4_0 {z(z^1^/^2-z^-^1^/^2)} \, dz[/tex] [distribute]
[tex]\int\limits^4_0 {z^3^/^2-z^1^/^2} \, dz[/tex] [split into 2 integrals]
[tex]\int\limits^4_0 {z^3^/^2} \, dz -\int\limits^4_0 {z^1^/^2} \, dxz[/tex] [solve integral for each part]
[tex]\frac{64}{5} -\frac{16}{3}[/tex] [solve]
[tex]\frac{112}{15}[/tex]
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D. *Note: I can't put -1 for the interval, but know that the 1 on the bottom is supposed to be -1.
[tex]\int\limits^1_1 {(1+u)(1-u)} \, du[/tex] [expand]
[tex]\int\limits^1_1 {1-u^2} \, du[/tex] [split into 2 integrals]
[tex]\int\limits^1_1 {1} \, du-\int\limits^1_1 {u^2} \, du[/tex] [solve integral for each part]
[tex]2-\frac{2}{3}[/tex] [solve]
[tex]\frac{4}{3}[/tex]
