$\int_{a}^{b} {(z + v)}^{n} d z$ $int_a^b(z+v)^ndz$

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Jul 24, 2020, 6:28:07 PM

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Jan 16, 2014, 6:07:14 AM

\int_{a}^{b} {(z + v)}^{n} d z = \frac{{(b + v)}^{n + 1}}{n + 1} - \frac{{(a + v)}^{n + 1}}{n + 1} + 0.0

$int_a^b(z+v)^ndz = ( "b" + "v" )^( "n" +1) / ( "n" +1) - ( "a" + "v" )^( "n" +1) / ( "n" +1) + 0.0$

(a) Lower Bound

$(a)"Lower Bound"$

(b) Upper Bound

$(b)"Upper Bound"$

(n) Integer Exponent

$(n)"Integer Exponent"$

(v) constant

$(v)"constant"$

(C) constant

$(C)"constant"$

∫ba(z+v)ndzint_a^b(z+v)^ndz