tolong jawab soal no 4 a-d saya beri nilai 80

Penyelesaian:

a. [tex] \int\limits^a_0 {2x + 3} \, dx = 4[/tex]
= x² + 3x ] 0->a = 4
= (a² + 3a) - (0) = 4
= a² + 3a - 4 = 0
(a+4) v (a-1)
a = -4 (TIDAK MEMENUHI)
a = 1 (MEMENUHI)

b. [tex] \int\limits^a_0 {3x^2} \, dx = 8 [/tex]
= x³ ] 0->a = 8
= a³ - (0) = 8
a³ = 8
a = 2

c. [tex] \int\limits^a_0 { \frac{1}{3}x^2 } \, dx = 4[/tex]
= 1/9.x³ ] 0->a = 4
= (1/9.a³) - (0) = 4
a³ = 36
a = [tex] \sqrt[3]{36} [/tex]

d. [tex] \int\limits^8_a { \frac{1}{5}x^1/3 } \, dx = \frac{9}{4} [/tex]
= 3/20.x^4/3 ] 0->8 = 9/4
= ((3/20).(8)^4/3) - ((3/20.a^4/3) = 9/4
= (3/20.16) - ((3/20.a^4/3) = 9/4
= 12/5 - ((3/20.a^4/3) = 9/4
= -3/20.a^4/3 = -3/20
a^4/3 = 1
a = 1