Question

1)Найти первообразную функции
2)Вычислить интеграл
3) Найти площадь фигуры ограниченной линиями

Answer 1

$f(x)=sqrt{x}-{2oversqrt{x}}\ F(x)=int(sqrt{x}-{2oversqrt{x}}){mathrm dx}=2{sqrt{x^3}over3}-4sqrt{x}+C\ f(x)={2over x^3}-{4over x^2}\ F(x)=int({2over x^3}-{4over x^2}){mathrm dx}=-x^{-2}+4x^{-1}+C\ f(x)=sqrt{6x-2}\ F(x)=intsqrt{6x-2}{mathrm dx}={1over6}intsqrt{6x-2}{mathrm d(6x-2)}={2over6*3}sqrt{(6x-2)^3}={1over9}sqrt{(6x-2)^3}$$\ int_{1}^{4}{sqrt{x}over x}{mathrm dx}=int_{1}^{4}(x^{{1over2}-1}){mathrm dx}=int_{1}^{4}x^{-{1over2}}{mathrm dx}=2sqrt{x}|_{1}^{4}=2(2-1)=2\ int_{-2}^{0}(x^5-3x^2){mathrm dx}={x^6over6}|_{-2}^{0}-x^3|_{-2}^{0}={1over6}(0-64)-0-8=-{32over3}-{24over3}=-{56over3}$
$\ int_{0}^{2}(sqrt{3-y}-{yover2}){mathrm dy}=-int_{0}^{2}sqrt{3-y}{mathrm d(3-y)}-{1over2}int_{0}^{2}y{mathrm dy}=-{2over3}(3-y)^{3over2}|_{0}^{2}-{1over4}y^2|_{0}^{2}=-{2over3}(1-3sqrt3)-1=2sqrt{3}-{5over3}$
4. f(x)=1-4x
F(x)=∫(1-4x)dx=x-2x^2+C
M(-1;9)
9= -1-2+C
C=12
F(x)=x-2x^2+12