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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;\](https://tex.z-dn.net/?f=f%28x%29%3D2x%2B4%3B%5C%0AF%28x%29-%3F%5C%0AB%28-1%3B1%29in+F%28x%29%5C%0AF%28x%29%3Dint%7Bf%28x%29%7Ddx%3Dint%7B%282x%2B4%29%7Ddx%3D2int%7Bx%7Ddx%2B4int%7B%7Ddx%3D%5C%0A%3D2cdotfrac%7Bx%5E%7B1%2B1%7D%7D%7B1%2B1%7D%2B4frac%7Bx%5E%7B0%2B1%7D%7D%7B0%2B1%7D%2BC%3Dx%5E2%2B4x%2BC%3B%5C%0Ax%3D-1%3B+y%3D1%3B%5C%0A1%3D%28-1%29%5E2%2B4%28-1%29%2BC%3D1-4%2BC%3B%5C%0AC%3D4%3B%5C%0AC%3D4%3B%5C%0AF%28x%29%3Dx%5E2%2B4x%2B4%3B%5C%0A "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;\](https://tex.z-dn.net/?f=intlimits_1%5E9%7Bfrac%7B6%7D%7Bsqrt+x%7D%7Ddx%3Dintlimits_1%5E9%7B6cdot+x%5E%7B-frac%7B1%7D%7B2%7D%7D%7D%7Ddx%3D6intlimits_1%5E9%7Bx%5E%7B-frac%7B1%7D%7B2%7D%7D%7D%7Ddx%3D6cdotfrac%7B1%7D%7B-frac%7B1%7D%7B2%7D%2B1%7Dx%5E%7B-frac%7B1%7D%7B2%7D%2B1%7D%7C_1%5E9%3D12sqrt%7Bx%7D%7C_1%5E9%3D%5C%0A%3D12%28sqrt9-sqrt1%29%3D12%283-1%29%3D12cdot2%3D24%3B%5C%0A "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_{-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;\](https://tex.z-dn.net/?f=intlimits_%7B-5%7D%5E1%7B%28x%5E2%2B8x%2B16%29%7Ddx%3Dfrac%7Bx%5E3%7D%7B3%7D%7C_%7B-5%7D%5E1%2B4x%5E2%7C_%7B-5%7D%5E1%2B16x%7C_%7B-5%7D%5E1%3D%5C%0A%3Dfrac%7B1%5E3-%28-5%29%5E3%7D%7B3%7D%2B4%281%5E2-%28-5%29%5E2%29%2B16%281-%28-5%29%29%3D%5C%0A%3Dfrac%7B1%2B125%7D%7B3%7D%2B4%281-25%29%2B16%281%2B5%29%3D%5C%0A%3Dfrac%7B126%7D%7B3%7D%2B4cdot%28-24%29%2B16cdot6%3D%5C%0A%3D42-96%2B96%3D42%3B%5C%0A%0A "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](https://tex.z-dn.net/?f=S%3Dintlimits_1%5E2%7B%282x%5E2-2%29%7Ddx%3Dfrac%7B2x%5E3%7D%7B3%7D%7C_1%5E2-2x%7C_1%5E2%3Dfrac%7B2%282%5E3-1%5E3%29%7D%7B3%7D-2%282-1%29%3D%5C%0A%3Dfrac%7B2%288-1%29%7D%7B3%7D-2cdot1%3Dfrac%7B2cdot7%7D%7B3%7D-2%3Dfrac%7B14%7D%7B3%7D-frac%7B6%7D%7B3%7D%3Dfrac83%3D2frac23 "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")
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