jika pada integral batas bawah a batas atas 3 (2×+1)dx:10 nilai a

(2x + 1) dx = 10
x² + x = 10
(3² + 3) - (a² - a) = 10
11 - a² + a = 10
0 = a² - a + 10 - 11
a² - a - 1 = 0
a = {-(-1) ± √(-1)² - 4(1)(-1)} / 2(1)
a = {1 ± √5} / 2
a = (1+√5) / 2
atau
a = (1-√5)/2