integral of-(t-6)^2

解答

\int - (t - 6)^{2} d t

- \frac{t ^{3}}{3} + 6 t^{2} - 36 t + C

求解步骤

\int - (t - 6)^{2} d t

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int (t - 6)^{2} d t

乘开

(t - 6)^{2} : t^{2} - 12 t + 36

= - \int t^{2} - 12 t + 36 d t

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int t^{2} d t - \int 12 t d t + \int 36 d t)

\int t^{2} d t = \frac{t ^{3}}{3}

\int 12 t d t = 6 t^{2}

\int 36 d t = 36 t

= - (\frac{t ^{3}}{3} - 6 t^{2} + 36 t)

化简

= - \frac{t ^{3}}{3} + 6 t^{2} - 36 t

解答补常数

= - \frac{t ^{3}}{3} + 6 t^{2} - 36 t + C

\frac{d}{d y} (x^{0.5} y^{0.5})

d er i v a t i v e x^{\frac{7}{9}}

t an g e n t f (x) = x^{2} - 7, at x = 2

x \to 2 - lim (\frac{x ^{2}}{x ^{2} + 4})

\frac{d}{d θ} (1 + 2 sin (θ))