Tentukan nilai dari integral 1 -2 (x pangkat 2 - 4x + 2) dx

integraltertentu

₋₂¹ ∫ x² - 4x + 2 dx =

= 1/3 x³ - 2x² + 2x]¹₋₂
= 1/3 (1³ -(-2)³)  - 2(1² - (-2)²) +  2(1 -(-2))
= 1/3 (1+8)  - 2 (1- 4) + 2 (1 + 2)
= 1/3 (9) - 2(-3) + 2(3)
= 3 +6 + 6
= 15

Materi integral

 \int\limits^1_{-2}\ {x^2-4x+2} \, dx = [(1/3)x^3-2x^2+2x] dengan batas 1 sampai -2
Berarti,
(1/3)1^3 - 2(1)^2 + 2(1) - ((1/3)(-2)^3 -2(-2)^2 + 2(-2))
1/3 - 2 + 2 - (-8/3 - 8 - 4)
= 1/3 - (-8/3 - 36/3) = 1/3 + 44/3 = 15