Soal yang Akan Dibahas
$ \int \limits_\frac{1}{2}^1 \left( \sqrt[3]{2x-1} + \sin \pi x \right) \, dx = .... $
A). $ \frac{3\pi - 8}{8\pi} $
B). $ \frac{3\pi - 4}{4\pi} $
C). $ \frac{3\pi + 4}{4\pi} $
D). $ \frac{3\pi + 8}{8\pi} $
E). $ \frac{3}{4} + \pi $
A). $ \frac{3\pi - 8}{8\pi} $
B). $ \frac{3\pi - 4}{4\pi} $
C). $ \frac{3\pi + 4}{4\pi} $
D). $ \frac{3\pi + 8}{8\pi} $
E). $ \frac{3}{4} + \pi $
$\spadesuit $ Konsep Dasar Integral
$ \int (ax+b)^n \, dx = \frac{1}{a(n+1)} (ax+b)^{n+1} + c $
$ \int \sin ax \, dx = -\frac{1}{a} \cos ax + c $
$ \int (ax+b)^n \, dx = \frac{1}{a(n+1)} (ax+b)^{n+1} + c $
$ \int \sin ax \, dx = -\frac{1}{a} \cos ax + c $
$\clubsuit $ Pembahasan
*). Menentukan Integralnya :
$\begin{align} & \int \limits_\frac{1}{2}^1 \left( \sqrt[3]{2x-1} + \sin \pi x \right) \, dx \\ & = \int \limits_\frac{1}{2}^1 \left( (2x-1)^\frac{1}{3} + \sin \pi x \right) \, dx \\ & = \left[ \frac{1}{2} . \frac{3}{4} (2x-1)^\frac{4}{3} - \frac{1}{\pi} \cos \pi x \right]_\frac{1}{2}^1 \\ & = \left[ \frac{3}{8} (2x-1)^\frac{4}{3} - \frac{1}{\pi} \cos \pi x \right]_\frac{1}{2}^1 \\ & = \left[ \frac{3}{8} (2.1-1)^\frac{4}{3} - \frac{1}{\pi} \cos \pi .1 \right] \\ & \, \, \, \, \, - \left[ \frac{3}{8} (2.\frac{1}{2}-1)^\frac{4}{3} - \frac{1}{\pi} \cos \pi . \frac{1}{2} \right] \\ & = \left( \frac{3}{8} + \frac{1}{\pi} \right) - \left( 0 - 0 \right) \\ & = \frac{3}{8} + \frac{1}{\pi} \\ & = \frac{3\pi + 8}{8\pi} \end{align} $
Jadi, hasil integralnya adalah $ \frac{3\pi + 8}{8\pi} . \, \heartsuit $
*). Menentukan Integralnya :
$\begin{align} & \int \limits_\frac{1}{2}^1 \left( \sqrt[3]{2x-1} + \sin \pi x \right) \, dx \\ & = \int \limits_\frac{1}{2}^1 \left( (2x-1)^\frac{1}{3} + \sin \pi x \right) \, dx \\ & = \left[ \frac{1}{2} . \frac{3}{4} (2x-1)^\frac{4}{3} - \frac{1}{\pi} \cos \pi x \right]_\frac{1}{2}^1 \\ & = \left[ \frac{3}{8} (2x-1)^\frac{4}{3} - \frac{1}{\pi} \cos \pi x \right]_\frac{1}{2}^1 \\ & = \left[ \frac{3}{8} (2.1-1)^\frac{4}{3} - \frac{1}{\pi} \cos \pi .1 \right] \\ & \, \, \, \, \, - \left[ \frac{3}{8} (2.\frac{1}{2}-1)^\frac{4}{3} - \frac{1}{\pi} \cos \pi . \frac{1}{2} \right] \\ & = \left( \frac{3}{8} + \frac{1}{\pi} \right) - \left( 0 - 0 \right) \\ & = \frac{3}{8} + \frac{1}{\pi} \\ & = \frac{3\pi + 8}{8\pi} \end{align} $
Jadi, hasil integralnya adalah $ \frac{3\pi + 8}{8\pi} . \, \heartsuit $
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