How do you integrate #\int _ { 1} ^ { 3} x ^ { 3} - 4x ^ { 2} + 3x d x#?

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Answer 1 · 2018-01-07T22:16:24

#-8/3#

#int_1^3 (x^3-4x^2+3x)*dx#

=#[1/4x^4-4/3x^3+3/2x^2]_1^3#

=#1/4*80-4/3*26+3/2*8#

=#-8/3#

Answer 2 · 2018-01-07T22:27:37

#int_1^3""x^3-4x^2+3x# #dx=-8/3#

Given: #int_1^3""x^3-4x^2+3x# #dx# we will solve the indefinite integral and then plug in the upper/lower limits but we can rewrite the integral as:

#intx^3dx-int4x^2dx+int3xdx#

We can integrate each integral separately to simplify the problem

#intx^3dx=x^4/4#

#int-4x^2dx->-4intx^2dx=-4{x^3/3}=-(4x^3)/3#

#int3xdx->3intxdx=3{x^2/2}=(3x^2)/2#

Thus,

#intx^3-4x^2+3x=x^4/4-(4x^3)/3+(3x^2)/2+"C"#

Now we evaluate the integral:

#[x^4/4-(4x^3)/3+(3x^2)/2]_1^3#

#={((3)^4)/4-(4(3)^3)/3+(3(3)^2)/2}-{((1)^4)/4-(4(1)^3)/3+(3(1)^2)/2}#

#=[-9/4]-[5/12]=-8/3#