Question

60 баллов...................

Answer 1

$1)\; \int _0^1(x^2+x)^2dx=\int _0^1(x^4+2x^3+x^2)dx=(\frac{x^5}{5}+\frac{2x^4}{4}+\frac{x^3}{3})|_0^1=\\\\=\frac{1}{5}+\frac{1}{2}+\frac{1}{3}+\frac{6+15+10}{30}=\frac{31}{30}\\\\2)\; \int _0^{\frac{\pi}{2}}(cos^2\frac{x}{2}-sin^2\frac{x}{2})dx=\int _0^{\frac{\pi}{2}}cosx\, dx=sinx|_0^{\frac{\pi}{2}}=\\\\=sin\frac{\pi}{2}-sin0=1\\\\3)\; \int_3^5|x|\, dx=[\, 3 \leq x \leq 5\; \to \; |x|=x\, ]=\\\\=\int _3^5x\, dx=\frac{x^2}{2}|_3^5=\frac{1}{2}(25-9)=8$

Answer 2

$\int\limits^1_0 {(x^2+x)^2} \, dx = \int\limits^1_0 {(x^4+2x^3+x^2)} \, dx = \frac{x^5}{5}+ \frac{x^4}{2}+ \frac{x^3}{3}|^1_0= \frac{31}{30}$

$\int\limits^{ \frac{\pi}{2} }_0 {(\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}) } \, dx = \int\limits^{\frac{\pi}{2}}_0 {\cos x} \, dx =\sin x|^{\frac{\pi}{2}}_0=1$

$\int\limits^5_3 {|x|} \, dx =5+3=8$