Question

Just making sure my answers are right. Please show the steps clearly. Thank you so much

Answer 1

Answer: See below

Step-by-step explanation:

A.

Let's split the integral into two parts, by the Sum Rule.

$\int\limits {x-4x^3} \, dx$                              [split into 2 integrals]

$\int\limits {x} \, dx -\int\limits {4x^3} \, dx$                      [solve integral for each part]

$\frac{1}{2} x^2-x^4+C$                            [Remember, we need to add C for constant]

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B.

$\int\limits {\frac{1+x}{\sqrt{x} } } \, dx$                                    [expand into 2 integrals]

$\int\limits {\frac{1}{\sqrt{x} } } \, dx +\int\limits {\frac{x}{\sqrt{x} } } \, dx$                    [simplify second integral]

$\int\limits {\frac{1}{\sqrt{x} } } \, dx +\int\limits {\sqrt{x} } \, dx$                    [solve integral for each part]

$2\sqrt{x} +\frac{2}{3}x^3^/^2+C$

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C.

$\int\limits^4_0 {z(z^1^/^2-z^-^1^/^2)} \, dz$                [distribute]

$\int\limits^4_0 {z^3^/^2-z^1^/^2} \, dz$                       [split into 2 integrals]

$\int\limits^4_0 {z^3^/^2} \, dz -\int\limits^4_0 {z^1^/^2} \, dxz$              [solve integral for each part]

$\frac{64}{5} -\frac{16}{3}$                                     [solve]

$\frac{112}{15}$

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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.

$\int\limits^1_1 {(1+u)(1-u)} \, du$                  [expand]

$\int\limits^1_1 {1-u^2} \, du$                              [split into 2 integrals]

$\int\limits^1_1 {1} \, du-\int\limits^1_1 {u^2} \, du$                      [solve integral for each part]

$2-\frac{2}{3}$                                      [solve]

$\frac{4}{3}$