Całki oznaczone

[true_direction]

1)\(\displaystyle{ \int_{-2}^{-1} \frac{1}{ (11 + 5x)^{3} } dx}\)
2)\(\displaystyle{ \int_{0}^{2a} \frac{3}{2b - x}}\) ,0

[soku11]

1.
\(\displaystyle{ \int\limits_{-2}^{-1} \frac{\mbox{d}x}{ (11 + 5x)^3 }=
ft\{\begin{array}{c}
11+5x=t\\
5\mbox{d}x=\mbox{d}t\\
=\frac{1}{5}\mbox{d}t\\
x=-1\;\Rightarrow\; t=6\\
x=-2\;\Rightarrow\; t=1
\end{array}\right\}=
\frac{1}{5} t\limits_{1}^{6} \frac{\mbox{d}t}{t^3}=
-\frac{1}{10} ft(\frac{1}{t^2}\right)\; ft|\frac{}{}\right|_{1}^{6}=
-\frac{1}{10} ft(\frac{1}{36}-1\right)=
\frac{7}{72}}\)

[ Dodano: 29 Listopada 2008, 18:50 ]
2.
\(\displaystyle{ 3\int\limits_{0}^{2a} \frac{\mbox{d}x}{2b - x} =
ft\{\begin{array}{c}
2b-x=t\\
-\mbox{d}x=\mbox{d}t\\
=-\mbox{d}t\\
x=0\;\Rightarrow\; t=2b\\
x=2a\;\Rightarrow\; t=2b-2a
\end{array}\right\}=
-3\int\limits_{2b}^{2b-2a} \frac{\mbox{d}t}{t}=
-3 (\ln t)\;\left|\frac{}{}\right|_{2b}^{2b-2a}=
-3[ \ln(2b-2a)-\ln (2b) ]=
-3[\ln 2+\ln (b-a)-\ln 2 -\ln b ]=
-3[\ln (b-a)-\ln b ]=
-3\ln \frac{b-a}{b}=
3\ln \frac{b}{b-a}}\)

[ Dodano: 29 Listopada 2008, 18:55 ]
3)
\(\displaystyle{ \int\limits_{0}^{ \frac{\pi}{2} } \sin^3 x =
t\limits_{0}^{ \frac{\pi}{2} } \sin x\cdot \sin^2 x =
t\limits_{0}^{ \frac{\pi}{2} } \sin x(1-\cos^2 x) =
ft\{\begin{array}{c}
\cos x=t\\
-\sin x\mbox{d}x=\mbox{d}t\\
\sin x\mbox{d}x=-\mbox{d}t\\
x=0\;\Rightarrow\; t=1\\
x=\frac{\pi}{2}\;\Rightarrow\; t=0
\end{array}\right\}=
-\int\limits_{1}^{0}(1-t^2) \mbox{d}t=
t\limits_{1}^{0}(t^2-1)\mbox{d}t=
ft[ \frac{t^3}{3}-t\right]\; ft|\frac{}{}\right_{1}^{0}=
-\frac{1}{3}+1=\frac{2}{3}}\)

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