Question

please bantuin .....​

Answer 1

Verified answer

6). (1/2x^(3 + 1))/(3 + 1) + C
(1/2x^4)/4 + C
1/2x^4 x 1/4 + C
1/8x^4 + C.

7). (-1/2x^(3 + 1))/(3 + 1) + C
(-1/2x^4)/4 + C
-1/2x^4 x 1/4 + C
-1/8x^4 + C.

8). (-1/2x^(-3 + 1))/(-3 + 1) + C
(-1/2x^(-2))/(-2) + C
-1/2x^(-2) x (-1/2) + C
1/4x^(-2) + C.

Semoga membantu.

Answer 2

Nomor 6

[tex]\begin{aligned}\displaystyle\tt~\int~ \frac{1}{2} {x}^{3} ~dx & = \displaystyle\tt~ \frac{ \frac{1}{2} }{3 + 1} {x}^{3 + 1} + C \\ \\ & = \displaystyle\tt~ \frac{ \frac{1}{2} }{4} {x}^{4} + C \\ \\ & = \displaystyle\tt~ \frac{1}{8} {x}^{4} + C \end{aligned}[/tex]

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Nomor 7

[tex]\begin{aligned}\displaystyle\tt~\int~ - \frac{1}{2} {x}^{3} ~dx & = \displaystyle\tt~ \frac{ - \frac{1}{2} }{3 + 1} {x}^{3 + 1} + C \\ \\ & = \displaystyle\tt~ \frac{ - \frac{1}{2} }{4} {x}^{4} + C \\ \\ & = \displaystyle\tt~ - \frac{1}{8} {x}^{4} + C \end{aligned}[/tex]

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Nomor 8

[tex]\begin{aligned}\displaystyle\tt~\int~ - \frac{1}{2} {x}^{ - 3}~dx & = \displaystyle\tt~ \frac{ - \frac{1}{2} }{ - 3 + 1} {x}^{ - 3 + 1} + C \\ \\ & = \displaystyle\tt~ \frac{ - \frac{1}{2} }{ - 2} {x}^{ - 2} + C \\ \\ & = \displaystyle\tt~ \frac{1}{4} {x}^{ - 2} + C \end{aligned}[/tex]