Question

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

Answer 1

`int_(v_1)^(v_3) (b-av)dv=b(v_3-v_1)-a/2(v_3^2-v_1^2)`

= `(v_3-v_1)(b-a/2(v_3+v_1))`

Explanation:

`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))`

Answer 2

See explanation.

Explanation:

`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)`

Answer 3

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

Explanation:

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))`