Find the value of integral $\int_{v_{1}}^{v_{3}} (b - a v) d v$ $int_(v_1)^(v_3) (b-av)dv$ ?

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Find the value of integral $\int_{v_{1}}^{v_{3}} (b - a v) d v$ $int_(v_1)^(v_3) (b-av)dv$ ?

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Answer 1 · 2017-06-20T16:45:11

$\int_{v_{1}}^{v_{3}} (b - a v) d v = b (v_{3} - v_{1}) - \frac{a}{2} (v_{3}^{2} - v_{1}^{2})$ $int_(v_1)^(v_3) (b-av)dv=b(v_3-v_1)-a/2(v_3^2-v_1^2)$

= $(v_{3} - v_{1}) (b - \frac{a}{2} (v_{3} + v_{1}))$ $(v_3-v_1)(b-a/2(v_3+v_1))$

Explanation:

$\int_{v_{1}}^{v_{3}} (b - a v) d v$ $int_(v_1)^(v_3) (b-av)dv$

= $\int_{v_{1}}^{v_{3}} b d v - \int_{v_{1}}^{v_{3}} a v d v$ $int_(v_1)^(v_3) bdv-int_(v_1)^(v_3) avdv$

= $b \int_{v_{1}}^{v_{3}} d v - a \int_{v_{1}}^{v_{3}} v d v$ $bint_(v_1)^(v_3) dv-aint_(v_1)^(v_3) vdv$

= $b {[v]}_{v_{1}}^{v_{3}} - a {[\frac{1}{2} v^{2}]}_{v_{1}}^{2}$ $b[v]_(v_1)^(v_3)-a[1/2v^2]_(v_1)^(v_3)$

= $b (v_{3} - v_{1}) - \frac{a}{2} (v_{3}^{2} - v_{1}^{2})$ $b(v_3-v_1)-a/2(v_3^2-v_1^2)$

or $(v_{3} - v_{1}) (b - \frac{a}{2} (v_{3} + v_{1}))$ $(v_3-v_1)(b-a/2(v_3+v_1))$

Answer 2 · 2017-06-20T16:45:13

See explanation.

Explanation:

$\int_{v_{2}}^{v_{3}} (b - a v) d v =$ $int_{v_2}^{v_3} (b-av) dv=$

${[b v - \frac{a v^{2}}{2}]}_{v = v_{2}}^{v = v_{3}}$ $[bv-(av^2)/2]_{v=v_2}^{v=v_3}$

$= b v_{3} - {(a v_{3})}_{3}^{2} - (b v_{2} - {(a v_{2})}_{2}^{2}) =$ $=bv_3-(av_3)^2-(bv_2-(av_2)^2)=$

$b v_{3} - a^{2} v_{3}^{2} - b v_{2} + a^{2} v_{2}^{2} =$ $bv_3-a^2v_3^2-bv_2+a^2v_2^2=$

$b (v_{3} - v_{2}) - a^{2} (v_{3}^{2} - v_{2}^{2}) =$ $b(v_3-v_2)-a^2(v_3^2-v_2^2)=$

$b (v_{3} - v_{2}) - a^{2} (v_{3} - v_{2}) (v_{3} + v_{2}) =$ $b(v_3-v_2)-a^2(v_3-v_2)(v_3+v_2)=$

$(v_{3} - v_{2}) \cdot (b - a^{2} (v_{3} + v_{2})) =$ $(v_3-v_2)*(b-a^2(v_3+v_2))=$

$(v_{3} - v_{2}) \cdot (b - a^{2} v_{3} - a^{2} v_{2}))$ $(v_3-v_2)*(b-a^2v_3-a^2v_2))$

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

$\int (a) d x = a x$ $int(a)dx=ax$
$\int (x^{n}) d x = \frac{x^{n + 1}}{n + 1}$ $int (x^n)dx=(x^(n+1))/(n+1)$

Answer 3 · 2017-06-20T16:49:04

$(v_{3} - v_{2}) (b - \frac{a}{2} (v_{3} + v_{2}))$ $(v_3 -v_2)(b- a/2(v_3 + v_2))$

Explanation:

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

$\int_{v_{2}}^{v_{3}} b d v - \int_{v_{2}}^{v_{3}} a v d v$ $int_(v_2)^(v_3)bdv - int_(v_2)^(v_3)avdv$

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

$b \int_{v_{2}}^{v_{3}} d v - a \int_{v_{2}}^{v_{3}} v d v$ $bint_(v_2)^(v_3)dv - aint_(v_2)^(v_3)vdv$

Tables of integrals should give you the integration of these two

${[b v - \frac{a}{2} v^{2}]}_{v_{2}}^{2}$ $[bv - a/2v^2]_(v_2)^(v_3)$

Evaluate the integral at the two limits of integration

$[b v_{3} - \frac{a}{2} {(v_{3})}_{3}^{2}] - [b v_{2} - \frac{a}{2} {(v_{2})}_{2}^{2}]$ $[bv_3 - a/2(v_3)^2]-[bv_2 - a/2(v_2)^2]$

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

$b v_{3} - \frac{a}{2} {(v_{3})}_{3}^{2} - b v_{2} + \frac{a}{2} {(v_{2})}_{2}^{2}$ $bv_3 - a/2(v_3)^2-bv_2 + a/2(v_2)^2$

$b v_{3} - b v_{2} - \frac{a}{2} {(v_{3})}_{3}^{2} + \frac{a}{2} {(v_{2})}_{2}^{2}$ $bv_3 -bv_2- a/2(v_3)^2 + a/2(v_2)^2$

$b (v_{3} - v_{2}) - \frac{a}{2} ({(v_{3})}_{3}^{2} - {(v_{2})}_{2}^{2})$ $b(v_3 -v_2)- a/2((v_3)^2 - (v_2)^2)$

$b (v_{3} - v_{2}) - \frac{a}{2} (v_{3} - v_{2}) (v_{3} + v_{2})$ $b(v_3 -v_2)- a/2(v_3 - v_2)(v_3 + v_2)$

$(v_{3} - v_{2}) (b - \frac{a}{2} (v_{3} + v_{2}))$ $(v_3 -v_2)(b- a/2(v_3 + v_2))$

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