Question

Помогите нужно решить срочно

Answer 1

1) $f(x) = F'(x)\\ 1-sinx+3e^{3x} = (x+cosx + e^{3x})' \\ 1-sinx+3e^{3x} = x'+cosx'+(e^{3x})' \\ 1-sinx+3e^{3x} = 1-sinx+3(e^{3x})$ чтд 
2)  $f(x) = -3 \sqrt[3]{x} \\ F(x) = \int\limits { -3 \sqrt[3]{x}} \, dx = \int\limits {-3 x^{\frac{1}{3}} } \, dx = \frac{-3}{ \frac{1}{3}+1} x^{\frac{1}{3}+1} = \frac{-9}{4} x^{\frac{4}{3}} + C \\ \frac{3}{4} = 0 \cdot \frac{-9}{4}+C \\ C=\frac{3}{4} \\ F(x) = -\frac{9}{4} x^{\frac{4}{3}} + \frac{3}{4}$ 
3) $\int\limits^3_2 { -x^2+6x-5} \, dx = |^{3}_{2} = -\frac{x^3}{3}+3x^2-5x = \frac{11}{3}$ 
4)  $\int\limits^3_1 { x^2 + \frac{3}{x}} \, dx = |^{3}_{1} = \frac{x^3}{3} + 3lnx = \frac{26}{3}+3ln3 \\ \int\limits^ {\frac{\pi}{2}}_0 { \sin^2x} \, dx = |^{\frac{\pi}{2}}_{0} \frac{2x-sin2x}{4} = \frac{\pi}{4}$ 
5) 
 $3-2x = x^2+3x-3 \\ x^2+5x-6=0 \\ (x+6)(x-1) = 0 \\ x=-6\\ x=1\\\ \int\limits^1_{-6} { (x^2+3x-3-3+2x)} \, dx = |^{1}_{-6} \frac{x^3}{3} + \frac{5x^2}{2}-6x = |\frac{343}{6}$