Question

Обчислити визначений інтеграл за допомогою формули
Ньютона-Лейбніця.

Answer 1

а)
$\displaystyle \int\limits^3_0 {\frac{dx}{x^2-25} } =\int\limits^3_0 {\frac{dx}{x^2-5^2} } =\frac{1}{2*5}ln|\frac{x-5}{x+5} | |^3_0=\frac{1}{10}*(ln|\frac{3-5}{3+5}| -ln|\frac{0-5}{0+5} | )=\\=0,1*(ln|-\frac{2}{8} |-ln|-\frac{5}{5} |)=0,1*(ln(\frac{1}{4})-ln(1) )=0,1*(ln(2^{-2})-0)=\\=-0,2*ln(2)$

б)
$\displaystyle \int\limits^7_1 {\frac{1}{1+\sqrt{9x+1} } } \, dx =$
Пусть 9х+1 = у, тогда 9dx = dy ⇔ dx = dy/9
$\displaystyle =\int\limits^7_1 {\frac{1}{9}* \frac{1}{1+\sqrt{y} } } \, dy =$
Пусть 1+√y = t(√y = t-1), тогда dy/(2√y) = dt ⇔ dy = (2√y)dt ⇒ dy = 2(t-1)dt
$\displaystyle =\frac{1}{9}*\int\limits^7_0 {\frac{2(t-1)}{t} } \, dt=\frac{2}{9}*(\int\limits^7_0 {\frac{t}{t} } \, dt-\int\limits^7_0 {\frac{1}{t} } \, dt)=\frac{2}{9}*(t|^7_0-ln|t|^7_0 )=$
Вернёмся к замене
$\displaystyle =\frac{2}{9}*((1+\sqrt{9x+1} )|^7_0-ln|1+\sqrt{9x+1}||^7_0)=\frac{2}{9}*(1+\sqrt{9*7+1}-(1+\sqrt{9*0+1} )-(ln|1+\sqrt{9*7+1}|-ln|1+\sqrt{9*0+1}|) ) =\frac{2}{9}*(1+\sqrt{64}-1-\sqrt{1}-\\-(ln|1+\sqrt{64} |-ln|1+\sqrt{1} |))=\frac{2}{9}*(8-1-(ln|1+8|-ln|1+1|))=\frac{2}{9}*(7-(ln(9)-ln(2))) =\frac{2}{9}*(7-ln(9)+ln(2))$

в)
$\displaystyle \int\limits^{1/2}_0 {\frac{\sqrt{arcsin(x)} }{\sqrt{1-x^2} } } \, dx =$
Пусть arcsin(x) = y, тогда dx/√(1-x²) = dy ⇔ dx = √(1-x²)dy
$\displaystyle \int\limits^{1/2}_0 {\sqrt{y} } \, dy =\int\limits^{1/2}_0 {y^{1/2}} \, dy = \frac{2y^{3/2}}{3}| ^{1/2}_0=$
г)
$\displaystyle \int\limits^e_1 {\frac{ln(x)}{x^4} } \, dx =\int\limits^e_1 {ln(x)*x^{-4}} \, dx =$
Пусть ln(x) = u, тогда dx/x = du
Пусть x⁻⁴dx = dV ⇔ -1/(3x⁻³) = V
$\displaystyle =ln(x)*(-\frac{1}{3x^3})|^e_1-\int\limits^e_1 {-\frac{1}{3x^3}*\frac{1}{x} } \, dx =-\frac{1}{3}*\frac{ln(x)}{x^3}|^e_1+\frac{1}{3}\int\limits^e_1 {\frac{1}{x^4} } \, dx =\\=-\frac{1}{3}*(\frac{ln(e)}{e^3}-\frac{ln(1)}{1^3} ) +\frac{1}{3}*(-\frac{1}{3x^3} )|^e_1=-\frac{1}{3}*\frac{1}{e^3}-\frac{1}{9}*(\frac{1}{e^3}-\frac{1}{1^3} ) =\\=-\frac{1}{3e^3}-\frac{1}{9e^3}+\frac{1}{9}$