Kak tolong di jawab pakai cara nomor 2 sampai 4

Caranya di gambar yaa

2.)

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

[tex]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}[/tex]

3.

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

[tex]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}[/tex]

4.

Kurva [tex]y = 2-x[/tex] (berdasarkan titik potong).
Interval [tex]2 \leq x \leq 3[/tex].

[tex]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}[/tex]
*) Diberi negatif, karena daerah berada di bawah sumbu X.