1.3 KiB
\lim_{ n \to \infty } \sum_{i=1}^{n} f(\xi_{i})∂x_{i}=\int^{b}_{a} f(x)dx=F(b)-F(a)=F(x)|^b_{a}
\lambda(P)\to 0
\lambda(P)=max(\Delta x_{i})
C+F(x)=\int f(x)dx
\int^{b}_{a}(\alpha f+\beta g)(x)dx=\alpha \int^b_{a}f(x)dz+\beta \int^b_{a}g(x)dx
\int^b_{a}f(x) \, dx=\int ^c_{a} f(x)\, dx+\int ^b_{c}f(x) \, dx; \forall a,b,c
\int_{a}^{b} f(x) \, dx=-\int_{b}^{a} f(x) \, dx
\int_{a}^{b} (uv )'x\, dx=(uv)(x)|^b_{a}-\int_{a}^{b} (u'v)(x) \, dx
\int_{a}^{b} f(x) \, dx=\int_{\alpha}^{\beta} f(\gamma(t))\phi'(t) \, dt; \gamma(\alpha)=a,\gamma(\beta)=b
Пример 1
#семестр_1 #высшая_математика
\int_{-1}^{1} \sqrt{ 1-x^2 } \, dx=
$$\begin{array} \
x=\sin t \
dx=\cos t\cdot dt \
\sin (2)=1; \alpha=\arcsin1=\frac{\pi}{2}
\sin(\beta)=-1; \beta=-\frac{\pi}{2}
\end{array}$$
=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{ 1-\sin^2t }\cos t \, dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2t \, dt=\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1+\cos 2t) \, dt=
=\frac{1}{2}t|^ \frac{\pi}{2}_{-\frac{\pi}{2}}-\frac{1}{4}\sin 2t|^ \frac{\pi}{2}_{\frac{\pi}{2}}=\frac{1}{2}\left( \frac{\pi}{2}+\frac{\pi}{2} \right)-\frac{1}{4}(0-0)=\frac{\pi}{2}
\int_{-1}^{7} \frac{dt}{\sqrt{ 3t+4 }}
(3t+4)^{-\frac{1}{2}}
\frac{1}{5}-\dots