How do I evaluate #int_1^4sqrt(t) \(5 + 2 t) dt#?

Question

1760 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-01-28T00:44:00

I would expand the argument as:

#int_1^4sqrt(t)*5+sqrt(t)*2tdt=#

I can now separate the integrals, as:

#int_1^4sqrt(t)*5dt+int_1^4sqrt(t)*2tdt=#

Remembering that #sqrt(t)=t^(1/2)# and the rules to manipulate exponentes, to get:

#=int_1^4t^(1/2)*5dt+int_1^4t^(1/2+1)*2dt=#
#=int_1^4t^(1/2)*5dt+int_1^4t^(3/2)*2dt=#

Taking the constant out of the integrals and remembering that the integral of #x^n# is #x^(n+1)/(n+1)#;

#5t^(1/2+1)/(1/2+1)+2t^(3/2+1)/(3/2+1)|_1^4#
#10/3t^(3/2)+4/5t^(5/2)|_1^4=48,13#