Pembahasan Soal Matematika Integral Kalkulus
Oke sahabatku semua disini saya ingin membahas soal matematika di Whatsapp saya, buat kalian yang suka dengan artikel kami silahkan klik berlangganan di bawah postingan ini. Oke semoga kalian suka😍😉
1) a.
$\int_{200}^{\infty} e^x\ dx$
$=\lim\limits_{a\to\infty}\int_{200}^{a} e^x\ dx$
$=\lim\limits_{a\to\infty}(e^a-e^{200})$
$=\infty-e^{200}$
$=\infty$
Integral tersebut divergen
b.
$\int_{-\infty}^{\infty} \frac{x}{\sqrt{x^2+25}}\ dx$
$=\lim\limits_{a\to -\infty}\int_{a}^{0} \frac{x}{\sqrt{x^2+25}}\ dx$ $+\lim\limits_{b\to\infty}\int_{0}^{b} \frac{x}{\sqrt{x^2+25}}\ dx$
Substitusi $x=5tan\ u$
$dx=5sec^2\ u\ du$
$=\lim\limits_{a\to -\infty}\int_{a}^{0} \frac{5tan\ u}{\sqrt{25tan^2\ u+25}}\ 5sec^2\ u\ du$ $+\lim\limits_{b\to\infty}\int_{0}^{b} \frac{5tan\ u}{\sqrt{25tan^2\ u+25}}\ 5sec^2\ u\ du$
$=\lim\limits_{a\to -\infty}\int_{a}^{0} \frac{5tan\ u}{5sec\ u}\ 5sec^2\ u\ du$ $+\lim\limits_{b\to\infty}\int_{0}^{b} \frac{5tan\ u}{5sec\ u}\ 5sec^2\ u\ du$
$=\lim\limits_{a\to -\infty}\int_{a}^{0} 25tan\ u.sec\ u\ du$ $+\lim\limits_{b\to\infty}\int_{0}^{b} 25tan\ u.sec\ u\ du$
$=\lim\limits_{a\to -\infty}(25sec\ u)$ $+\lim\limits_{b\to\infty} (25sec\ u)$
$=\lim\limits_{a\to -\infty}(25-5\sqrt{a^2+25})$ $+\lim\limits_{b\to\infty} (5\sqrt{b^2+25}-25)$
$=\lim\limits_{a\to \infty}(-5\sqrt{a^2+25})$ $+\lim\limits_{b\to\infty} (5\sqrt{b^2+25})$
$=\lim\limits_{a\to\infty}\int_{200}^{a} e^x\ dx$
$=\lim\limits_{a\to\infty}(e^a-e^{200})$
$=\infty-e^{200}$
$=\infty$
Integral tersebut divergen
b.
$\int_{-\infty}^{\infty} \frac{x}{\sqrt{x^2+25}}\ dx$
$=\lim\limits_{a\to -\infty}\int_{a}^{0} \frac{x}{\sqrt{x^2+25}}\ dx$ $+\lim\limits_{b\to\infty}\int_{0}^{b} \frac{x}{\sqrt{x^2+25}}\ dx$
Substitusi $x=5tan\ u$
$dx=5sec^2\ u\ du$
$=\lim\limits_{a\to -\infty}\int_{a}^{0} \frac{5tan\ u}{\sqrt{25tan^2\ u+25}}\ 5sec^2\ u\ du$ $+\lim\limits_{b\to\infty}\int_{0}^{b} \frac{5tan\ u}{\sqrt{25tan^2\ u+25}}\ 5sec^2\ u\ du$
$=\lim\limits_{a\to -\infty}\int_{a}^{0} \frac{5tan\ u}{5sec\ u}\ 5sec^2\ u\ du$ $+\lim\limits_{b\to\infty}\int_{0}^{b} \frac{5tan\ u}{5sec\ u}\ 5sec^2\ u\ du$
$=\lim\limits_{a\to -\infty}\int_{a}^{0} 25tan\ u.sec\ u\ du$ $+\lim\limits_{b\to\infty}\int_{0}^{b} 25tan\ u.sec\ u\ du$
$=\lim\limits_{a\to -\infty}(25sec\ u)$ $+\lim\limits_{b\to\infty} (25sec\ u)$
$=\lim\limits_{a\to -\infty}(25-5\sqrt{a^2+25})$ $+\lim\limits_{b\to\infty} (5\sqrt{b^2+25}-25)$
$=\lim\limits_{a\to \infty}(-5\sqrt{a^2+25})$ $+\lim\limits_{b\to\infty} (5\sqrt{b^2+25})$
$=0$
Integral tersebut Konvergen
2) a.
Titik potong grafik
$y+1=3-y^2$
$y^2+y-2=0$
$(y-1)(y+2)=0$
$y=1$ atau $y=-2$
Cari koordinat
$y=1\rightarrow x=2$
$y=-2\rightarrow x=-1$
$(2,1)$ dan $(-1,2)$
Luas daerah
$\int_{-2}^1 \left((3-y^2)-(y+1)\right)\ dy$
$=\int_{-2}^1 \left(-y^2-y+2\right)\ dy$
$=-\frac{1}{3}y^3-\frac{1}{2}y^2+2y|_{-2}^1$
$=-\frac{1}{3}-\frac{1}{2}+2-\frac{8}{3}+2+4$
$=\frac{5}{2}$
b.
Titik potong grafik
$y=0\rightarrow x=2$
$x=0\rightarrow y=4$
$(2,0)$ dan $(0,4)$
Luas daerah
$\int_{0}^2 \left(4-x^2\right)\ dx$
$=4x-\frac{1}{3}x^3|_{0}^2$
$=8-\frac{8}{3}-0$
$=\frac{16}{3}$
Integral tersebut Konvergen
2) a.
Titik potong grafik
$y+1=3-y^2$
$y^2+y-2=0$
$(y-1)(y+2)=0$
$y=1$ atau $y=-2$
Cari koordinat
$y=1\rightarrow x=2$
$y=-2\rightarrow x=-1$
$(2,1)$ dan $(-1,2)$
Luas daerah
$\int_{-2}^1 \left((3-y^2)-(y+1)\right)\ dy$
$=\int_{-2}^1 \left(-y^2-y+2\right)\ dy$
$=-\frac{1}{3}y^3-\frac{1}{2}y^2+2y|_{-2}^1$
$=-\frac{1}{3}-\frac{1}{2}+2-\frac{8}{3}+2+4$
$=\frac{5}{2}$
b.
Titik potong grafik
$y=0\rightarrow x=2$
$x=0\rightarrow y=4$
$(2,0)$ dan $(0,4)$
Luas daerah
$\int_{0}^2 \left(4-x^2\right)\ dx$
$=4x-\frac{1}{3}x^3|_{0}^2$
$=8-\frac{8}{3}-0$
$=\frac{16}{3}$
Oke guys pembahasannya cukup sampai di sini, kalian juga bisa melihat pembahasan soal integral lainnya di blog ini.
