n-ый член арифметической прогрессии ищется по формуле:
![b_n=b_1+(n-1)d](https://tex.z-dn.net/?f=b_n%3Db_1%2B%28n-1%29d)
Рассмотрим восемнадцатый член:
![b_{18}=b_1+17d~~~\Leftrightarrow~~~ d=\dfrac{b_{18}-b_1}{17}=\dfrac{-8-2}{17}=-\dfrac{10}{17}](https://tex.z-dn.net/?f=b_%7B18%7D%3Db_1%2B17d~~~%5CLeftrightarrow~~~%20d%3D%5Cdfrac%7Bb_%7B18%7D-b_1%7D%7B17%7D%3D%5Cdfrac%7B-8-2%7D%7B17%7D%3D-%5Cdfrac%7B10%7D%7B17%7D)
<span>у=log₂₀(x²-x)
</span>log₂₀(x²-x)>0
x²-x>1
x²-x-1 = (x-1/2+√5/2)(x-√5/2-1/2)>0
x∈(-∞, 1/2-√5/2)⋃(√5/2+1/2, ∞)
4. Под знак дифференциала постепенно загоняем: сначала косинус, затем двойку и наконец единицу, т.е.
![\frac{1}{2}d(2sinx+1) =cosxdx](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B2%7Dd%282sinx%2B1%29+%3Dcosxdx)
, чтобы получился табличный интеграл от степенной функции.
![\int\limits^{ \pi /2}_0 { \sqrt{2sinx+1} * cosx} \, dx =\int\limits^{ \pi /2}_0 { \sqrt{2sinx+1} } \, d(sinx) = \\ \\ =\int\limits^{ \pi /2}_0 { \frac{1}{2} \sqrt{2sinx+1} } \, d(2sinx+1) = \frac{1}{2}\int\limits^{ \pi /2}_0 {(2sinx+1)^{ \frac{1}{2} } } \, d(2sinx+1) = \\ \\ = \frac{1}{2} \frac{2}{3} (2sinx+1)^{ \frac{3}{2}}= \frac{1}{3} (2sinx+1)^{ \frac{3}{2}}|_{0}^{\pi /2}= \\ \\ = \frac{1}{3} (2sin \frac{ \pi }{2} +1)^{ \frac{3}{2}} -\frac{1}{3} (2sin 0 +1)^{ \frac{3}{2}} =](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E%7B+%5Cpi+%2F2%7D_0+%7B+%5Csqrt%7B2sinx%2B1%7D+%2A+cosx%7D+%5C%2C+dx+%3D%5Cint%5Climits%5E%7B+%5Cpi+%2F2%7D_0+%7B+%5Csqrt%7B2sinx%2B1%7D+%7D+%5C%2C+d%28sinx%29+%3D+%5C%5C++%5C%5C+%3D%5Cint%5Climits%5E%7B+%5Cpi+%2F2%7D_0+%7B+%5Cfrac%7B1%7D%7B2%7D+%5Csqrt%7B2sinx%2B1%7D+%7D+%5C%2C+d%282sinx%2B1%29+%3D+%5Cfrac%7B1%7D%7B2%7D%5Cint%5Climits%5E%7B+%5Cpi+%2F2%7D_0+%7B%282sinx%2B1%29%5E%7B+%5Cfrac%7B1%7D%7B2%7D+%7D+%7D+%5C%2C+d%282sinx%2B1%29+%3D+%5C%5C++%5C%5C+%3D+%5Cfrac%7B1%7D%7B2%7D++%5Cfrac%7B2%7D%7B3%7D+%282sinx%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D%3D+%5Cfrac%7B1%7D%7B3%7D+%282sinx%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D%7C_%7B0%7D%5E%7B%5Cpi+%2F2%7D%3D+%5C%5C++%5C%5C+%3D+%5Cfrac%7B1%7D%7B3%7D+%282sin+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+-%5Cfrac%7B1%7D%7B3%7D+%282sin+0+%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+%3D)
![= \frac{1}{3} (2sin \frac{ \pi }{2} +1)^{ \frac{3}{2}} -\frac{1}{3} (2sin 0 +1)^{ \frac{3}{2}} = \frac{1}{3} \sqrt{27} -\frac{1}{3}= \sqrt{3} -\frac{1}{3}](https://tex.z-dn.net/?f=%3D+%5Cfrac%7B1%7D%7B3%7D+%282sin+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+-%5Cfrac%7B1%7D%7B3%7D+%282sin+0+%2B1%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+%3D+%5Cfrac%7B1%7D%7B3%7D+%5Csqrt%7B27%7D+-%5Cfrac%7B1%7D%7B3%7D%3D+%5Csqrt%7B3%7D++-%5Cfrac%7B1%7D%7B3%7D)
5.
![\int\limits^4_2 { \frac{1}{x-1} } \, dx =\int\limits^4_2 { \frac{1}{x-1} } \, d(x-1) =ln(x-1)|_{2}^{4}=ln3-ln1=ln3](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E4_2+%7B+%5Cfrac%7B1%7D%7Bx-1%7D+%7D+%5C%2C+dx+%3D%5Cint%5Climits%5E4_2+%7B+%5Cfrac%7B1%7D%7Bx-1%7D+%7D+%5C%2C+d%28x-1%29+%3Dln%28x-1%29%7C_%7B2%7D%5E%7B4%7D%3Dln3-ln1%3Dln3)