Question

∫ batas atas 3 , batas bawah 1 (3x²-3x + 2) dx

tolong dibantu ya kak soal integral !!​

Answer 1

Penjelasan dengan langkah-langkah:

[tex]\tt \int\limits^3_1 {(3x^2-3x+2)} \, dx\\\\\\= 3\int\limits^3_1 {x^2} \, dx-3x\int\limits^3_1 {x} \, dx+2x\\ \\\\=\frac{3x^{2+1}}{2+1}-\frac{3x^{1+1}}{1+1}+2\\ \\ = x^3-\frac{3x^2}{2} +2x|^3_1\\\\= 3^3-\frac{3(3)^2}{2}+2(3)-(1^3-\frac{3(1)^2}{2}+2(1))\\ \\= 27-\frac{27}{2}+6-(1-\frac{3}{2}+2)\\ \\ = 33-\frac{27}{2}-\frac{3}{2}\\ \\ = 18[/tex]