Integral batas atas 2 batas bawah 0, 4x (x^2-2)^4

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Integral batas atas 2 batas bawah 0, 4x (x^2-2)^4

Jawaban Terkonfirmasi

displaystyle text{misal:}\x^2-2=u\2x,dx=du\\intlimits_0^24x(x^2-2)^4,dx=intlimits_0^2u^4cdot4x,dx\intlimits_0^24x(x^2-2)^4,dx=intlimits_0^2u^4cdot2,du\intlimits_0^24x(x^2-2)^4,dx=intlimits_0^22u^4,du\intlimits_0^24x(x^2-2)^4,dx=leftfrac25u^5right|_0^2\intlimits_0^24x(x^2-2)^4,dx=leftfrac25(x^2-2)^5right|_0^2\intlimits_0^24x(x^2-2)^4,dx=frac25(2^2-2)^5-frac25(0^2-2)^5\boxed{boxed{intlimits_0^24x(x^2-2)^4,dx=25frac35}}

Integral Substitusi
∫ 4x (x² – 2)⁴ dx
Misal u = x² – 2
du / dx = 2x (turunan u)
dx = du / 2x
∫ 4x u⁴ dx
∫ 4x u⁴ du /2x
∫ 2 u⁴ du
Hasil integral
2/5 u⁵ batas [0 – 2]
[2/5 ((2)² – 2)⁵] – [2/5((0)² -2)⁵]
[2/5(2)⁵] – [2/5(-2)⁵]
2/5.(32) – 2/5(-32)
64/5 – (-64/5)
128 / 5
Jadi hasil integral = 128 / 5

Ok?