Find the value of integral #int_(v_1)^(v_3) (b-av)dv#?

#int_(v_1)^(v_3) (b-av)dv#

= #int_(v_1)^(v_3) bdv-int_(v_1)^(v_3) avdv#

= #bint_(v_1)^(v_3) dv-aint_(v_1)^(v_3) vdv#

= #b[v]_(v_1)^(v_3)-a[1/2v^2]_(v_1)^(v_3)#

= #b(v_3-v_1)-a/2(v_3^2-v_1^2)#

or #(v_3-v_1)(b-a/2(v_3+v_1))#

#int_{v_2}^{v_3} (b-av) dv=#

#[bv-(av^2)/2]_{v=v_2}^{v=v_3}#

#=bv_3-(av_3)^2-(bv_2-(av_2)^2)=#

#bv_3-a^2v_3^2-bv_2+a^2v_2^2=#

#b(v_3-v_2)-a^2(v_3^2-v_2^2)=#

#b(v_3-v_2)-a^2(v_3-v_2)(v_3+v_2)=#

#(v_3-v_2)*(b-a^2(v_3+v_2))=#

#(v_3-v_2)*(b-a^2v_3-a^2v_2))#

In the first line I used the following properties of integral:

• #int(a)dx=ax#

• #int (x^n)dx=(x^(n+1))/(n+1)#

When you integrate to expressions subtracted from each other, integrate each item separately.

#int_(v_2)^(v_3)bdv - int_(v_2)^(v_3)avdv#

You can bring the constants of integration out to the front.

#bint_(v_2)^(v_3)dv - aint_(v_2)^(v_3)vdv#

Tables of integrals should give you the integration of these two

#[bv - a/2v^2]_(v_2)^(v_3)#

Evaluate the integral at the two limits of integration

#[bv_3 - a/2(v_3)^2]-[bv_2 - a/2(v_2)^2]#

This is really a good stopping point, but you can try to simplify if you like as well.

#bv_3 - a/2(v_3)^2-bv_2 + a/2(v_2)^2#

#bv_3 -bv_2- a/2(v_3)^2 + a/2(v_2)^2#

#b(v_3 -v_2)- a/2((v_3)^2 - (v_2)^2)#

#b(v_3 -v_2)- a/2(v_3 - v_2)(v_3 + v_2)#

#(v_3 -v_2)(b- a/2(v_3 + v_2))#