Kak tolong di jawab pakai cara nomor 2 sampai 4.... Question from @Rita06121

msandyka 2.)

Kurva $y = x^{2} + 3$ .
Interval $1 \leq x \leq 3$ .
Tentukan luas arsiran terhadap sumbu X.

$L = \int\limits^a_b {y} \, dx \\
L = \int\limits^3_1(x^{2}+3)dx \\
L = [\frac{1}{2+1}x^{2+1}+3x]^3_1 \\
L = [\frac{1}{3}x^{3}+3x]^3_1 \\
L = [\frac{1}{3}(3)^{3}+3(3)] - [\frac{1}{3}(1)^{3}+3(1)] \\
L = [9+9]-[\frac{1}{3}+3] \\
L = 18 - 3\frac{1}{3} \\
L = 14\frac{2}{3}$

3.

Kurva $y = x^{3}+1$ .
Interval $2 \leq x \leq 3$ .
Tentukan luas arsiran terhadap sumbu X.

$L = \int\limits^3_2 (x^{3}+1) dx \\
L = [\frac{1}{4}x^{4}+x]^3_2 \\
L = [\frac{1}{4}(3)^{4}+3]-[\frac{1}{4}(2)^{4}+(2)] \\
L = [\frac{81}{4}+\frac{12}{4}]-[\frac{16}{4}+\frac{8}{4}] \\
L = \frac{93}{4} - \frac{24}{4} \\
L = \frac{69}{4}$

4.

Kurva $y = 2-x$ (berdasarkan titik potong).
Interval $2 \leq x \leq 3$ .

$L = -\int\limits^3_2 (2-x) dx \\
L = -([2x-\frac{1}{2}x^{2}]^3_2) \\
L = -([6-\frac{9}{2}] - [4-2]) \\
L = -(\frac{3}{2}-2) \\
L = \frac{1}{2}$
*) Diberi negatif, karena daerah berada di bawah sumbu X.

1 votes Thanks 1

Kak tolong batu kerjakan pakai cara besok sudah di kumpulkan

Kak bisa jawab soal ini gak

Kak bisa jawab soal ini gak

Tolong di jawab ya kak

Tolong di jawab ya kak soal tentang integral, karang sudah di kumpulkan

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