Question

Și ăsta urgent!!Va rog!

Answer 1

$\displaystyle \mathtt{ a)~\int\limits^3_1 {\left(3x^2+2x+1\right)} \, dx =\int\limits^3_1 3x^2dx+\int\limits^3_12xdx+\int\limits^3_11dx=}\\ \\ \mathtt{=3\int\limits^3_1x^2dx+2\int\limits^3_1xdx+\int\limits^3_11dx=3 \frac{x^3}{3} \Bigg|^3_1+2 \frac{x^2}{2}\Bigg|^3_1+x\Bigg|^3_1=}\\ \\ \mathtt{=3\left( \frac{3^3}{3}- \frac{1^3}{3}\right)+2\left( \frac{3^2}{2}- \frac{1^2}{2} \right)+3-1= 3 \cdot \frac{28}{3}+2 \cdot \frac{8}{2}+2=}\\ \\ \mathtt{=26+8+2=34+2=36}$

$\displaystyle \mathtt{b)~ \int\limits^\pi_0(cos~x+sin~x)dx=\int\limits^\pi_0cos~x~dx+\int\limits^\pi_0sin~x~dx=}\\ \\ \mathtt{=sin~x \Bigg|^\pi_0-cos~x\Bigg|^\pi_0=(sin~\pi-sin~0)-(cos~\pi-cos~0)=}\\ \\ \mathtt{=(0-0)-(-1-1)=0-(-2)=0+2=2}$

$\displaystyle \mathtt{c)~ \int\limits^1_0xe^xdx}\\ \\ \mathtt{\int\limits f(x)g'(x)dx=f(x)g(x)-\int\limits f'(x)g(x)dx}\\ \\ \mathtt{f(x)=x\Rightarrow f'(x)=1}\\ \\ \mathtt{g'(x)=e^x\Rightarrow g(x)=\int\limits g'(x)dx=\int\limits e^xdx=e^x}\\ \\ \mathtt{\int\limits xe^xdx=xe^x-\int\limits e^xdx=xe^x-e^x+C=e^x(x-1)+C}\\ \\ \mathtt{\int\limits_0^1xe^xdx=e^x(x-1)\Bigg|^1_0=e^1(1-1)-e^0(0-1)=1}$

$\displaystyle \mathtt{d)~\int\limits_2^4 \frac{x^2-1}{x+1} dx=\int\limits_2^4 \frac{(x-1)(x+1)}{x+1}dx=\int\limits_2^4(x-1)dx=\int\limits_2^4xdx-\int\limits_2^41dx =}\\ \\ \mathtt{= \frac{x^2}{2}\Bigg|^4_2-x \Bigg|_2^4= \left( \frac{4^2}{2}- \frac{2^2}{2}\right)-(4-2)= \frac{12}{2}-2=6-2=4 }$

$\displaystyle \mathtt{e)~\int\limits_1^2(x+5)e^xdx}\\ \\ \mathtt{\int\limits f(x)g'(x)dx=f(x)g(x)-\int\limits f'(x)g(x)dx}\\ \\ \mathtt{f(x)=x+5 \Rightarrow f'(x)=1}\\ \\ \mathtt{g'(x)=e^x \Rightarrow g(x)=\int\limits g'(x)dx=\int\limits e^xdx=e^x}\\ \\ \mathtt{\int\limits (x+5)e^xdx=(x+5)e^x-\int\limits e^xdx=(x+5)e^x-e^x+C=}\\ \\ \mathtt{=e^x(x+4)+C}\\ \\ \mathtt{\int\limits_1^2(x+5)e^xdx=e^x(x+4)\Bigg|_1^2=e^2(2+4)-e^1(1+4)=6e^2-5e}$

$\displaystyle \mathtt{f)~\int\limits_1^3(2x+1)\left(x^2+x\right)dx=\int\limits_1^3\left(2x^3+3x^2+x\right)dx=}\\ \\ \mathtt{=\int\limits_1^32x^3dx+\int\limits_1^33x^2dx+\int\limits_1^3xdx=2\int\limits_1^3x^3dx+3\int\limits_1^3x^2dx+\int\limits_1^3xdx=}\\ \\ \mathtt{=2 \frac{x^4}{4}\Bigg|_1^3+3 \frac{x^3}{3}\Bigg|_1^3+ \frac{x^2}{2}\Bigg|_1^3=2\left( \frac{3^4}{4}- \frac{1^4}{4}\right)+3\left( \frac{3^3}{3}- \frac{1^3}{3}\right)+\left( \frac{3^2}{2}- \frac{1^2}{2}\right)=}\\ \\ \mathtt{=40+26+4=70}$

$\displaystyle \mathtt{g)~ \int\limits_0^3\left(2-4x+3x^2\right)dx= \int\limits_0^32dx- \int\limits_0^34xdx+ \int\limits_0^33x^2dx=}\\ \\ \mathtt{= \int\limits_0^32dx-4 \int\limits_0^3xdx+3 \int\limits_0^3x^2dx=2x\Bigg|_0^3-4 \frac{x^2}{2}\Bigg|_0^3+3 \frac{x^3}{3}\Bigg|_0^3= }\\ \\ \mathtt{=(2 \cdot3-2\cdot0)-4\left( \frac{3^2}{2}- \frac{0^2}{2}\right)+3\left( \frac{3^3}{3}- \frac{0^3}{3}\right)=6-18+27=15}$

$\displaystyle \mathtt{h)~\int\limits_1^4 \frac{1}{2 \sqrt{x} }dx= \frac{1}{2} \int\limits_1^4 \frac{1}{ \sqrt{x} }dx= \frac{1}{2} \int\limits_1^4\left( \sqrt{x} \right)^{-1}= \frac{1}{2}\int\limits_1^4x^{- \frac{1}{2} }= }\\ \\ \mathtt{= \frac{1}{2} \frac{x^{ \frac{1}{2} }}{ \frac{1}{2} } \Bigg|^4_1= \frac{1}{2} \left(\frac{4^{ \frac{1}{2} }}{ \frac{1}{2} }- \frac{1^{ \frac{1}{2} }}{ \frac{1}{2} } \right)= \frac{1}{2}\cdot \frac{2-1}{ \frac{1}{2} } = \frac{1}{2}\cdot 2=1 }$

$\displaystyle \mathtt{i)~ \int\limits _3^4\left(x^2-4x+3\right)dx=\int\limits _3^4x^2dx-\int\limits _3^44xdx+\int\limits _3^43dx=}\\ \\ \mathtt{=\int\limits _3^4x^2dx-4\int\limits _3^4xdx+\int\limits _3^43dx= \frac{x^3}{3}\Bigg|_3^4-4 \frac{x^2}{2}\Bigg|_3^4+3x\Bigg|_3^4=}\\ \\ \mathtt{=\left( \frac{4^3}{3}- \frac{3^3}{3}\right)-4\left( \frac{4^2}{2}- \frac{3^2}{2}\right)+(3 \cdot 4-3 \cdot 3)= \frac{37}{3} -14+3=}\\ \\ \mathtt{= \frac{37-42+9}{3}= \frac{4}{3} }$