Tentukan nilai dari
∫ -1 sampai 1 4x^6 - 4x^3 + 16x^2 - 4x + 8 dx![]()
∫ -1 sampai 1 4x^6 - 4x^3 + 16x^2 - 4x + 8 dx
Jawaban:
[tex] { ∫ _{ - 1}}^{1} (4 {x}^{6} - 4 {x}^{3} + 16 {x}^{2} - 4x + 8 )\: dx \\ {( \frac{4}{7} {x}^{7} - \frac{4}{4} {x}^{4} + \frac{16}{3} {x}^{3} - \frac{4}{2} {x}^{2} + 8x) _{ - 1}}^{1} \\ ( \frac{4}{7} \times 1 - 1 \times 1 + \frac{16}{3} \times 1 - 2 \times 1 + 8 \times 1) - ( \frac{4}{7} \times ( - 1) - 1 \times 1 + \frac{16}{3} \times ( - 1) - 2 \times 1 + 8 \times ( - 1)) \\ (\frac{4}{7} - 1 + \frac{16}{3} - 2 + 8) - ( - \frac{4}{7} - 1 - \frac{16}{3} - 2 - 8) \\ \frac{8}{7} + \frac{32}{3} + 16 \\ \frac{24 + 224 + 336}{21} \\ \frac{584}{21} \\ 27.8[/tex]
[answer.2.content]
