integral from 0 to t of (t-x)sin(x)

解答

\int_{0}^{t} (t - x) sin (x) d x

t - sin (t)

求解步骤

\int_{0}^{t} (t - x) sin (x) d x

乘开

(t - x) sin (x) : t sin (x) - x sin (x)

= \int_{0}^{t} t sin (x) - x sin (x) d x

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_{0}^{t} t sin (x) d x - \int_{0}^{t} x sin (x) d x

\int_{0}^{t} t sin (x) d x = t (- cos (t) + 1)

\int_{0}^{t} x sin (x) d x = sin (t) - t cos (t)

= t (- cos (t) + 1) - (sin (t) - t cos (t))

化简

t (- cos (t) + 1) - (sin (t) - t cos (t)) : t - sin (t)

= t - sin (t)

h \to + 0 - lim (\frac{2 ^{h} - 1}{h})

\int \frac{1}{x ^{2} - 8 x + 25} d x

\frac{d}{d t} (2 sin (3 t))

s l o p e (80, 2000), (260, 5600)

\int_{2}^{7} \frac{1}{25 + ( x - 2 ) ^{2}} d x