Ada yang bisa bantu ngerjain kalkulus?.... Question from @herawijaya

October 2018 1 38 Report

Ada yang bisa bantu ngerjain kalkulus?

Takamori37 Nomor 1.
$$\begin{align}\int_{-1}^3x\sqrt{x^2+2}\, dx&=\frac12\int_{-1}^3(x^2+2)^{1/2}(2x\, dx) \\ &=
\left.\frac12.\frac1{3/2}(x^2+2)^{3/2}\right]_{-1}^3\\&=
\left.\frac13(x^2+2)^{3/2}\right]_{-1}^3\\&=\frac13[(3^2+2)^{3/2}-((-1)^2+2)^{3/2}] \\ &=\frac13(11^{3/2}-3^{3/2}) \\ &\approx10,4289\end{align}$

Nomor 2.
Tentukan integralnya dengan parsial.
$$\begin{align}\int x^2.e^x\, dx&=x^2.e^x-\int 2x.e^x\, dx \\ &=x^2.e^x-(2x.e^x-\int2e^x\, dx) \\ &=x^2.e^x-2x.e^x+2e^x+C \\ &=e^x(x^2-2x+2)+C\end{align}$
Sehingga,
$$\begin{align}\int_{-2}^0x^2.e^x\, dx&=[e^0(0^2-2(0)+2)]-[e^{-2}((-2)^2-2(-2)+2)]\\&=1(2)-10e^{-2}\\&=2-10e^{-2} \\ &\approx0,64664\end{align}$

Nomor 3.
Sesuaikan dengan turunannya:
u = x²-2x+3
du = 2x-2 dx
Pisahkan:
$$\begin{align}\int\frac{2x+3}{x^2-2x+3}\, dx&=\int\frac{2x-2}{x^2-2x+3}\, dx+\int\frac5{x^2-2x+3}\, dx \\ &=\ln|x^2-2x+3|+\int\frac{5}{(x-1)^2+(\sqrt{2})^2}\, dx \\ &=\ln|x^2-2x+3|+\frac5{\sqrt{2}}\tan^{-1}\left[\frac{x-1}{\sqrt{2}}\right]+C\end{align}$
Substitusikan:
$$\begin{align}\int_{-2}^3\frac{2x+3}{x^2-2x+3}\, dx&=\left[\ln|9-6+3|+\frac5{\sqrt{2}}\tan^{-1}\left[\frac{3-1}{\sqrt{2}}\right]\right]\ddots \\ &~~~~-\left[\ln|4+4+3|+\frac5{\sqrt{2}}\tan^{-1}\left[\frac{-2-1}{\sqrt{2}}\right]\right] \\ &=\ln 6-\ln 11+\frac5{\sqrt{2}}\left[\tan^{-1}(\sqrt{2})-\tan^{-1}\frac{-3}{\sqrt{2}}\right] \\ &=-\ln\frac{11}6+\frac5{\sqrt{2}}\left[\tan^{-1}(\sqrt{2})+\tan^{-1}\frac{3}{\sqrt{2}}\right] \\ &\approx 6,7676\end{align}$

1 votes Thanks 1

herawijaya tanks bro, kerennnnnnnn mantappppsss

Takamori37 Oke :D