[+50](10/10)quiz terlalu umum di web

aplikasi integral

volume

cara 1

volume

= π ∫y² dx [4 1]

= π ∫(3x + 2)² dx

= π ∫(9x² + 12x + 4) dx

= π (3x³ + 6x² + 4x)

= π (3(4³ – 1³) + 6(4² – 1²) + 4(4 – 1))

= π (189 + 90 + 12)

= 291 satuan volume

•

cara 2

a = ∆x = 4 – 1 = 3

x = 4 → y = R = 3.4 + 2 = 14

x = 1 → y = r = 3.1 + 2 = 5

volume kerucut terpancung

= 1/3 πa (rR + r² + R²)

= 1/3 π . 3(5.14 + 5² + 14²)

= π (70 + 25 + 196)

= 291π satuan volume