Evaluar la integral

1) Evaluar la integral ∫_1^5▒e^x dx=e^x evaluado en e^x |_(1 )^5=e^x-e^1

=148.4131591-2.718281828=145.6948773

2) Calcular el área bajo la curva y=x^3 desde 0 a 1

∫_0^1▒〖x^3 dx = 〗 x^4/4 ]_(0 )^1=1^4/4-0^4/4= 1^ /4= .25 u^2

3) calcular ∫_0^2▒|2x-1|dx

de acuerdo a la definicion del numero absoluto tenemos que

|2x-1|= {■(2x-1 si 2x-1 ≥0@-(2x-1) si 2x-1 <0)

|2x-1|= {■(2x-1 si 2x≥1 x≥1/2@ -2x+1 si -2x < -1 x<1/2)

ahora se reescribe las integrales de acuerdo a lo que condiciona el valor absoluto

∫_0^(1/2)▒〖-2x+1 dx〗 + ∫_(1/2)^2▒〖2x-1 dx〗

resolviendo queda -(2x^2)/2+x |_(0 )^(1/2) + (2x^2)/2-x|_(1/2)^2 aplicando el TFC

= [-(2(1/2)^2)/2+1/2 ]-[-(2(0)^2)/2+0]+[ (2(2)^2)/2-2]-[ (2(1/2)^2)/2-1/2 ]

=[-(2(1/4)^ )/2+1/2 ]-[-0/2+0]+[ (2(4)^ )/2-2]-[ (2(1/4)^( ))/2-1/2 ]

=[-( 2^ )/8+1/2 ] +[ 8/2-2]-[ ( 2^( ))/( 8)-1/2 ]= [-( 2^ )/8+4/8 ] +[ 4-2]-[ ( 2^( ))/( 8)-( 4)/8 ]=5/2=2.5

4) hallar la integral de ∫_1^3▒dx/( x) =ln⁡〖x 〗 |_1^3 =ln⁡〖3-ln⁡〖1=ln⁡〖3/1〗 〗 〗=ln⁡3=1.0986

5) hallar la integral de (d )/dx [∫_0^x▒〖√(t^2+1) dt〗]

∫_0^x▒√(t^2+1) dt=t/2 √(t^2+1)+1/2 ln(t+√(t^2+1))|_0^x

(x/2 √(x^2+1)+1/2 ln(x+√(x^2+1)))-(t/2 √(0^2+1)+1/2 ln(0+√(0^2+1)))=

(x√(x^2+1)+ ln(x+√(x^2+1))-t)/2

6) evaluar la funcion F=∫_0^x▒cos⁡〖t dt 〗 x=0 ,π⁄6,π⁄4,π⁄3,π⁄2

∫_0^x▒cos⁡〖t dt 〗 = sen t |_0^0=0

∫_0^x▒cos⁡〖t dt 〗 = sen t |_0^(π⁄6)=sen π⁄6-sen 0= 0.5

∫_0^x▒cos⁡〖t dt 〗 = sen t |_0^(π⁄4)=sen π⁄4-sen 0= 0.707

∫_0^x▒cos⁡〖t dt 〗 = sen t |_0^(π⁄3)=sen π⁄3-sen 0= 0.866

∫_0^x▒cos⁡〖t dt 〗 = sen t |_0^(π⁄2)=sen π⁄2-sen 0= 1.0

7) evaluar la funcion F=∫_(π⁄2)^(x^3)▒cos⁡〖t dt 〗

∫_(π⁄2)^(x^3)▒cos⁡〖t dt 〗 = sen t |_(π⁄2)^(x^3 )=sen x^3-sen π⁄2

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