Hasil dari integral 0 sampe 1 X√1-X² dx adalah.... Question from @gracemayessi

adamdhijee $\int\limits^1_0 {x \sqrt{1-x^2} } \, dx \\\ u=1-x^2,\text{maka}~ \frac{du}{dx} =-2x ~\text{atau}~\text{x~dx}= -\frac{1}{2} ~\text{du}. \\\ \int\limits^1_0 {(1-x^2)^{ \frac{1}{2} }} \,x. dx \\\ \int\limits^1_0 {u^{ \frac{1}{2} }} \, .- \frac{1}{2}~du \\\ = \frac{1}{ \frac{1}{2}+1 } . -\frac{1}{2}u^{ \frac{1}{2} +1} ]^1_0 \\\ = \frac{1}{ \frac{3}{2} }.- \frac{1}{2} u^{ \frac{3}{2}} ]^1_0 \\\ =- \frac{1}{3} (1-x^2)^{ \frac{3}{2} } ]^1_0 \\\ =- \frac{1}{3} .(1-x^2). \sqrt{1-x^2} ]^1_0 \\\ =0-(-\frac{1}{3} )\Rightarrow \text{Masukan~angka~1~maka~hasilnya~tetap~0}\\\ =\frac{1}{3}$

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adamdhijee kalo ada yang ditanyakan silahkan.

gracemayessi ia ..makasihhh

Takamori37 Menggunakan integral substitusi:
$u=1-x^2 ~~\to~~du=-2x\, dx~~\to~~x\, dx=-\frac12\, du$
Sehingga:
$$\begin{align}\int x\sqrt{1-x^2}\, dx&=\int \sqrt u\times\left[-\frac12\, du\right] \\ &=-\frac12\int u^{1/2}\, du \\ &=-\frac12\times\frac1{3/2}u^{3/2}+C \\ &=-\frac13u^{3/2}+C \\ &=-\frac13(1-x^2)^{3/2}+C\end{align}$

Substitusikan:
$$\begin{align}\int_0^1x\sqrt{1-x^2}\, dx&=\left.-\frac13(1-x^2)^{3/2}\right|_0^1 \\ &=-\frac13[(1-1^2)^{3/2}-(1-0^2)^{3/2}] \\ &=-\frac13(0^{3/2}-1^{3/2}) \\ &=-\frac13(0-1) \\ &=-\frac13\times(-1) \\ &=\frac13\end{align}$

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Nilai dari lim x sampe 3 ( 2^x-9x+9/9-x² ) adalah..

Luas daerah yang d batasi oleh kurva y=2x-6 sumbu x, garis x=1,dan garis x=3 adalah

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