Question

Помогите с математикой, очень нужно, номера 2,3,4

Answer 1

2.
$f(x)=2x+4;\ F(x)-?\ B(-1;1)in F(x)\ F(x)=int{f(x)}dx=int{(2x+4)}dx=2int{x}dx+4int{}dx=\ =2cdotfrac{x^{1+1}}{1+1}+4frac{x^{0+1}}{0+1}+C=x^2+4x+C;\ x=-1; y=1;\ 1=(-1)^2+4(-1)+C=1-4+C;\ C=4;\ C=4;\ F(x)=x^2+4x+4;$

3.
$intlimits_1^9{frac{6}{sqrt x}}dx=intlimits_1^9{6cdot x^{-frac{1}{2}}}}dx=6intlimits_1^9{x^{-frac{1}{2}}}}dx=6cdotfrac{1}{-frac{1}{2}+1}x^{-frac{1}{2}+1}|_1^9=12sqrt{x}|_1^9=\ =12(sqrt9-sqrt1)=12(3-1)=12cdot2=24;$

$intlimits_0^fracpi2{cos x}dx=sin x|_0^fracpi2=sinfracpi2-sin0=1-0=1$

$intlimits_{-5}^1{(x^2+8x+16)}dx=frac{x^3}{3}|_{-5}^1+4x^2|_{-5}^1+16x|_{-5}^1=\ =frac{1^3-(-5)^3}{3}+4(1^2-(-5)^2)+16(1-(-5))=\ =frac{1+125}{3}+4(1-25)+16(1+5)=\ =frac{126}{3}+4cdot(-24)+16cdot6=\ =42-96+96=42;$

3.
$S=intlimits_1^2{(2x^2-2)}dx=frac{2x^3}{3}|_1^2-2x|_1^2=frac{2(2^3-1^3)}{3}-2(2-1)=\ =frac{2(8-1)}{3}-2cdot1=frac{2cdot7}{3}-2=frac{14}{3}-frac{6}{3}=frac83=2frac23$