4- Calcula les integrals següents: a) `\int_-1 ^1 (-x)/(x^2+1) dx` b) `\int_0 ^2 (2-3x)^5 dx` c) `\int_2^4 3/(1-x) dx` d) `\int_0 ^1 x/(1+x^4) dx` SOLUCIÓ:
`x^2+1=t => 2xdx=dt => xdx=dt/2` `\int_-1 ^1 (-x)/(x^2+1) dx=[-1/2ln(x^2+1)]_-1^1=-1/2[ln(1^2+1)-ln(1^2+1)]=-1/2[ln(2)-ln(2)]=0` `-1/2ln(4/2)=-1/2ln(2)=ln\sqrt{1/2}` b) `\int (2-3x)^5 dx=\int t^5 (-dt)/3=-1/3\int t^5 dt=-1/3t^6/6=(-t^6)/18=(-(2-3x)^6)/18` `(2-3x)=t => -3dx=dt => dx=-dt/3` `\int_0 ^2 (2-3x)^5 dx=[(-(2-3x)^6)/18]_0^2=(-(2-6)^6)/18-(-(2-0)^6)/18=-(-4)^6/18+2^6/18 = -224` c) `\int_2^4 3/(1-x) dx=[-3ln|1-x|]_2^4=-3[ln(3)-ln(1)]=-3ln(3)` Hi posem valor absolut, `ln|1-x|`, si no, no tindria sentit, ja que hi haurien logaritmes de nombres negatius. Això no ho demana l'exercici. d) `\int (x dx)/(1+x^4)=\int dt/(2(1+t^2))=1/2\int dt/(1+t^2)=1/2 arctan(t)=1/2arctan(x^2)` `x^2=t => 2x=dt => x=dt/2` `\int_0 ^1 x/(1+x^4) dx=1/2[arctan(x^2)]0_^1=1/2[arctan(1^2)-arctan(0^2)]=` `1/2[arctan(1)-arctan(0)]=1/2(\pi/4-0)=\pi/8` 5- Calcula: `\int_-2 ^1 x^2e^xdx`. Per trobar una primitiva cal aplicar el mètode d'integració per parts dues vegades. SOLUCIÓ:
`f'=e^x => f=e^x` `g=x^2 => g'=2x` `\int x^2e^xdx=x^2e^x -\int2xe^xdx =>x^2e^x -2\intxe^xdx`
`f'=e^x => f=e^x` `g=x => g'=1` `\intxe^xdx=xe^x-\int e^xdx=xe^x- e^x` `\int x^2e^xdx=x^2e^x -\int2xe^xdx =>x^2e^x -2\intxe^xdx=x^2e^x -2(xe^x- e^x)=(x^2 -2x+2)e^x` `\int_-2^1 x^2e^xdx=[(x^2 -2x+2)e^x]_-2^1=(1^2 -2·1+2)e^1-((-2)^2 -2(-2)+2)e^(-2)=e-10/e^2` 6- Troba una primitiva de la funció `x^3/(x-2)` i calcula la integral d'aquesta funció en l'interval `[3,5]`.
`\int x^3/(x-2)dx`
| 1 0 0 0 | 2 | 2 4 8 -----+--------------- 1 2 4 8 `x^3/(x-2)= x^2+2x+4 + 8/(x-2)` `\int x^3/(x-2)dx=\int (x^2+2x+4 + 8/(x-2))dx=x^3/3+x^2+4x-8ln|x-2|` `\int_3^5 x^3/(x-2)dx=[x^3/3+x^2+4x-8ln|x-2|]_3^5=` `(5^3/3+5^2+4·5-8ln|5-2|)-(3^3/3+3^2+4·3-8ln|3-2|)=` `(125/3+25+20-8ln(3))-(9+9+12-8ln(1))=` `(125/3+45-8ln|3|)-(30-8·0)=` `125/3+15-8ln(3)=170/3-8ln(3)` 7- Calcula `\int_0^2\sqrt{4-x^2}dx` utilitzant el canvi de variable: `x=2sin(t)` SOLUCIÓ:
`x=2sin(t) => dx=2cos(t)dt` `\int \sqrt{4-x^2}dx=\int \sqrt{4-4sin^2(t)}·2cos(t)dt=2\int \sqrt{4(1-sin^2(t))}·cos(t)dt=` `4\int \sqrt{1-sin^2(t)}·cos(t)dt=4\int cos(t)·cos(t)dt=4\int cos^2(t)dt=`*
`cos^2(t) = (1+cos(2t))/2` o per parts. Aquí teniu la demostració. Que és: `\int cos^2tdt=1/2(sin(t)cos(t)+t)` *`=4\int cos^2(t)dt=4·1/2(sin(t)cos(t)+t)=2[sin(t)cos(t)+t]=`** `x=2sin(t) => sin(t)=x/2 => cos(t)=\sqrt{1-sin^2(t)}=\sqrt{1-x^2/4}` i `t=arcsin(x/2)` **`2[sin(t)cos(t)+t]=2[x/2·\sqrt{1-x^2/4}+arcsin(x/2)]=` `2[x/2·\sqrt{4/4-x^2/4}+arcsin(x)]=2[x/4·\sqrt{4-x^2}+arcsin(x/2)]` Finalment recordar que el que estem intentant calcular és: `\int_0^2\sqrt{4-x^2}dx=2[(x/4·\sqrt{4-x^2}+arcsin(x/2))]_0^2=` `2[(2/4·\sqrt{4-2^2}+arcsin(2/2))-(0/4·\sqrt{4-0^2}+arcsin(0/2))]=` `2[(0+arcsin(1))-(0+arcsin(0))]=2[\pi/2-0]=\pi` |