How do you evaluate #\int_{1} ^{2}(3t^{2}-t)dt#?

2 Answers
Monzur R.
Nov 18, 2017

#int_1^2 (3 t^2 - t) dt = 11/2 #

Explanation:

We need to find #int_1^2 (3 t^2 - t) dt#. To this end, we shall use the reverse power rule of integration.

Reverse Power Rule

#intx^ndx=1/(n+1)x^(n+1) +"c"#

#thereforeint_1^2 (3 t^2 - t) dt=int_1^2 3 t^2dt - int_1^2t dt#

#=[3/3t^3-1/2t^2+"c"]_1^2=[t^3-1/2t^2+"c"]_1^2#

Here we don't need to include the constant as it would cancel out when we expand. However, I will add it for clarification.

#=(2^3-1/2*2^2+c)-(1^3-1/2*1^2+c)#

#=8+1-2+1/2+"c"-"c"=11/2#

Jim G.
Nov 18, 2017

#11/2#

Explanation:

#"integrate each term using the "color(blue)"power rule"#

#•color(white)(x)int(ax^n)=a/(n+1)x^(n+1)color(white)(x)n!=-1#

#rArrint_1^2(3t^2-t)dt#

#=[t^3-1/2t^2]_1^2#

#"to evaluate the integral"#

#int_a^bf(t)dt=[F(t]_a^b=F(b)-F(a)#

#rArr(8-2)-(1-1/2)=11/2#

Related questions
Impact of this question
1543 views around the world
You can reuse this answer
Creative Commons License