integral of |2x-3|

解答

\int ∣ 2 x - 3 ∣ d x

\frac{1}{4} ∣ 2 x - 3 ∣ (2 x - 3) + C

求解步骤

\int ∣ 2 x - 3 ∣ d x

使用换元积分法

= \int ∣ u ∣ \frac{1}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int ∣ u ∣ d u

使用积分常见定则:

\int ∣ u ∣ d u = \frac{u ∣ u ∣}{2}

= \frac{1}{2} \cdot \frac{u ∣ u ∣}{2}

u = 2 x - 3

代回

= \frac{1}{2} \cdot \frac{( 2 x - 3 ) ∣ 2 x - 3 ∣}{2}

化简

\frac{1}{2} \cdot \frac{( 2 x - 3 ) ∣ 2 x - 3 ∣}{2} : \frac{1}{4} ∣ 2 x - 3 ∣ (2 x - 3)

= \frac{1}{4} ∣ 2 x - 3 ∣ (2 x - 3)

解答补常数

= \frac{1}{4} ∣ 2 x - 3 ∣ (2 x - 3) + C

y^{^{''}} + 6 y^{^{'}} = - 108 cos (6 x)

\int \frac{( 5 sin ^{3} ( x ) )}{cos ( x )} d x

\int 5 e^{9 x} d x

t an g e n t f (x) = x^{4}, (2, 16)

x \to 0 lim (+ (x))