Question

Нужно решить 6 интегралов с объяснениями

Answer 1

$\displaystyle 1) \int\limits^1_0 {(3e^x-5*8^x+1)} \, dx = (3e^x-5*\frac{8^x}{ln8} +x) \ | ^1_0 = (3e^1-5*\frac{8^1}{ln8} +1)-\\\\ - (3e^0-5*\frac{8^0}{ln8} +0) = 3e-5*\frac{8}{ln8} +1-3*1-\frac{5*1}{ln 8} = 3e-2-\\\\-\frac{40-5}{ln 8} = 3e-2-\frac{35}{ln 8}$

$\displaystyle 2)\int\limits^4_{-4} {(10^{\frac{x}{4} }-sin\pi x)} \, dx = (\frac{1}{\frac{1}{4} }*\frac{10^{\frac{x}{4} }}{ln10} -\frac{1}{\pi} *(-cos\pi x)) \ |^4_{-4} = (\frac{4*10^{\frac{x}{4} }}{ln10} +\\\\ +\frac{cos \pi x}{\pi}) \ |^4_{-4} = (\frac{4*10^{\frac{4}{4} }}{ln10} +\frac{cos \ 4\pi }{\pi}) - (\frac{4*10^{\frac{-4}{4} }}{ln10} +\frac{cos -4\pi }{\pi})=$

$\displaystyle =\frac{4*10^1}{ln10} +\frac{1}{\pi} -\frac{4*10^{-1}}{ln10} -\frac{1}{\pi} =\frac{40}{ln10}-\frac{4*\frac{1}{10} }{ln10}= \frac{40-0,6}{ln10}= \frac{39,4}{ln10}$

$\displaystyle 3) \int\limits^{24}_6 {\frac{dx}{xln 2} } = \frac{1}{ln2}* ln|x| \ |^{24}_6 = \frac{ln24}{ln2} - \frac{ln6}{ln2} = \frac{ln24-ln6}{ln2} = \frac{ln(24:6)}{ln2} =\\\\ = \frac{ln2^2}{ln2} = 2$

$\displaystyle 4) \int\limits^{-2}_{-2,5} {\frac{8}{(2x+3)^3} } \, dx = 8*\int\limits^{-2}_{-2,5} {\((2x+3)^{-3}} \, dx = 8*\frac{1}{2} *\frac{(2x+3)^{-3+1}}{-3+1} \ |^{-2}_{-2,5} =\\\\ = 4*\frac{(2x+3)^{-2}}{-2} \ |^{-2}_{-2,5} = -\frac{2}{(2x+3)^2}\ |^{-2}_{-2,5} = -\frac{2}{(2*(-2)+3)^2}-\\\\ -(-\frac{2}{(2*(-2,5)+3)^2})=-\frac{2}{(-4+3)^2}+\frac{2}{(-5+3)^2} =-\frac{2}{1}+\frac{2}{4} = -1,5$

$\displaystyle 5) \int\limits^1_0 {(x-\sqrt{x} )^2} \, dx = \int\limits^1_0 {(x^2-2x^{\frac{3}{2} }+x) \, dx = (\frac{x^3}{3} -2*\frac{x^{\frac{3}{2}+1} }{{\frac{3}{2}+1}} +\frac{x^2}{2} ) | ^1_0=$

$\displaystyle = (\frac{x^3}{3}- \frac{4x^2\sqrt{x} }{5} +\frac{x^2}{2}) \ |^1_0 = (\frac{1^3}{3}- \frac{4*1^2\sqrt{1} }{5} +\frac{1^2}{2})- (\frac{0^3}{3}- \frac{4*0^2\sqrt{0} }{5} +\frac{0^2}{2}) = \\\\ =\frac{1}{3} - \frac{4*1*1}{5} + \frac{1}{2} =\frac{10}{30} -\frac{24}{30} + \frac{15}{30} =\frac{1}{30}$

$\displaystyle 6) \int\limits^{\frac{\pi}{6}} _0 {sin^4x} \, dx = \int\limits^{\frac{\pi}{6}} _0 \frac{1-cos^22x}{2} \ dx = \int\limits^{\frac{\pi}{6}} _0 (\frac{1}{2}-\frac{1+\frac{cos4x}{2} }{2}) \ dx = \int\limits^{\frac{\pi}{6}} _0(\frac{1}{2} -\frac{1}{2}+\\\\ +\frac{cos4x}{4} ) = \int\limits^{\frac{\pi}{6}} _0 \frac{cos 4x}{4} \ dx = \frac{1}{4}* \frac{1}{4} *sin4x \ | ^{\frac{\pi}{6}} _0 = \frac{sin(4*\frac{\pi}{6}) }{8} - \frac{sin(4*0) }{8}=$

$\displaystyle =\frac{sin\frac{2\pi}{3} }{8} -\frac{sin 0 }{8} = \frac{\sqrt{3} }{2} *\frac{1}{8} - 0 = \frac{\sqrt{3} }{16}$