[tex]\int\limits^1_0 {(x^4+1)} \, dx=(\frac{x^5}{5}+x)|_0^1=\frac{1^5}{5}+1-(\frac{0^5}{5}+0)=\frac{1}{5}+1-0=0,2+1=1,2[/tex]
[tex]\int\limits^1_0 {(x^3+3)} \, dx=(\frac{x^4}{4}+3x)|_0^1=\frac{1^4}{4}+3*1-(\frac{0^4}{4}+3*0)=\frac{1}{4}+3=0,75+3=3,75[/tex]
[tex]\int\limits^2_1 {x^2} \, dx=\frac{x^3}{3}|_1^2=\frac{2^3}{3}-\frac{1^3}{3}=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}=2\frac{1}{3}[/tex]
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[tex]\int\limits^1_0 {(x^4+1)} \, dx=(\frac{x^5}{5}+x)|_0^1=\frac{1^5}{5}+1-(\frac{0^5}{5}+0)=\frac{1}{5}+1-0=0,2+1=1,2[/tex]
[tex]\int\limits^1_0 {(x^3+3)} \, dx=(\frac{x^4}{4}+3x)|_0^1=\frac{1^4}{4}+3*1-(\frac{0^4}{4}+3*0)=\frac{1}{4}+3=0,75+3=3,75[/tex]
[tex]\int\limits^2_1 {x^2} \, dx=\frac{x^3}{3}|_1^2=\frac{2^3}{3}-\frac{1^3}{3}=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}=2\frac{1}{3}[/tex]