integral a sampai 1 (4×+3) dx = -3 nilai a adalah

[tex]\int_a^1 4x+3\, dx=-3 \\ 2x^2+x]_a^1=-3 \\ (2(1)^2+(1))-(2a^2+a)=-3 \\ 3-2a^2-a=-3 \\ 2a^2+a-6=0 \\ (a+2)(2a-3)=0 \\ a=-2$ dan $a=\frac{3}{2}[/tex]

[tex][tex]\int\limits^1_a {(4x+3)} \, dx =-3\\\\= \frac{4}{1+1}x^{1+1} + \frac{3}{0+1} x^{0+1} ] ^{1} _{a} =-3\\\\= 2x^{2} +3x]^1_a = -3\\\\=2(1)^2 + 3(1) - 2(a)^2+3(a) = -3\\\\ = (2+3) - 2a^2 +3a=-3\\\\2a^2 -3a - 8= 0\\\\(2a-7)(a+2) = -6\\\\a = \frac{1}{2} atau a = -8[/tex]