Integrales definidas

1.-∫_(-2)^3▒(6x^2-5) dx=∫_(-2)^3▒〖6x^2 dx〗-∫_(-2)^3▒〖5dx=〗 [(6x^3)/3]_(-2)^3-[5]_(-2)^3=[2x^3-5x]
=[2(3)^3-5(3) ]-[2(-2)^3-5(-2) ]=[39]+[6]=45
2.-∫_(-1)^2▒(x^3+1)^2dx=∫_(-1)^2▒(x^6+2x^3+1) dx=∫_(-1)^2▒x^6 dx+∫_(-1)^2▒〖2x^3 〗 dx∫_(-1)^2▒1 dx
〖 =[x^7/7]〗_(-1)^2 〖+2[x^4/4]〗_(-1)^2+[x]_(-1)^2=[x^7/7+x^4/2+x]= [128/7+16/2+2]-[(-1)/7+1/2-1]=[198/7]-[(-9)/14]=198/7+9/14=28.92
3.-∫_1^4▒〖(5x+2√x+32⁄x^3 )=〗 ∫_1^4▒〖5xdx+∫_1^4▒〖2x〗^(1⁄2) dx+∫_1^4▒32⁄x^3 〗 dx
〖 =[(5x^2)/2]〗_1^4+[〖4x/3〗^(3⁄2) ]_1^4+[16/x^2]_1^4=[(5〖(4)〗^2)/2+〖(4(4))/3〗^(3⁄2)+16/〖(4)〗^2 ]-[(5〖(1)〗^2)/2+〖(4(1))/3〗^(3⁄2)+16/〖(1)〗^2 ]
=[40+ 42.6+1]-[2.5+.66+16]=?
4.-∫_1^3▒(2x^3-4x^2+5)/x^2 dx=∫_1^3▒〖2x-4+5/x^2 =∫_1^3▒〖2xdx-∫_1^3▒4dx+∫_1^3▒〖5/x^2 dx〗〗〗
〖 =[x^2]〗_1^3-[4x]_1^3+[〖5x〗^(-1)/(-1)]_1^3=[x^2-4x-5/x]=[〖(3)〗^2-4(3)-5/((3))]-[〖(1)〗^2-4(1)-5/((1))]
=[98-12-5/3]-[-8]=[-4.66+8]=3.33
5.-∫_1^2▒〖(〖4u〗^(-5)+〖6u〗^(-4) )du=4∫_1^2▒〖u^(-5) du〗+6∫_1^2▒〖u^(-4)du=〖-u〗^(-4)-〗 2u^(-3) 〗=[^ -^ ]-
6.-∫_0^4▒〖x√(x^2+9)〗 dx= 1/2 ∫_0^4▒〖2x(x^2+9)^(1/2) 〗 dx=[1/3 (x^2+9)^(3/2) ]_0^4=1/3 [(25)^(3/2) 〖-9〗^(3/2) ]=98/3
7.-∫_2^3▒x/√(x^2-1)=1/2 ∫_2^3▒2x (x^2-1)^(-1/2)dx=[√(x^2-1)]_2^3=√8-√3=1.0963
8.-∫_(-2)^(-1)▒〖dx/(x-1)^3 =[(-1)/〖2(x-1)〗^2 ]_(-2 )^(-1)=1/2 [1/(-2)^2 -1/(-3)^2 ] 〗=-5/72
9.-∫_2^3▒〖dx/(x ln^4 x)=∫_2^3▒〖ln^(-4) x∙1/x dx=[-1/(3ln^3 x)]_2^3=-1/(3ln^33)+-1/(3ln^3 2)〗 〗

10.-∫_0^4▒dx/(1+√x)
x=〖〖t 〗^2 4=〖t 〗^2 0=〖t 〗^2〗^█(2 @ dx=2t dt) t=2 t=0
∫_0^4▒〖dx/(1+√x)=∫_0^2▒〖2t/(1+t)dt=2∫_0^2▒(1-1/(1+t))dt〗〗=2[t-ln(1+t) ]_0^2=4-2ln3

Integrales definidas

Dada una función f(x) de una variable real x y un intervalo [a,b] de la recta real, la integral definida es igual al área limitada entre la gráficade f(x), el eje de abscisas, y las líneas verticales x = a y x = b.

Se representa por.∫_a^b▒〖f(x)dx〗
∫ es el signo de integración.
a límite inferior de la integración.
b...