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Is this how a single integral works?

$int_b^a$ $(c)$ $dx$ $=$ $a$ $xx$ c $-$ $b$ $xx$ $c$

1 Answer

Jul 10, 2017

Assuming $c$ in the given equation is not dependent on the variable $x,$ LHS of given integral becomes

$int_b^a$ $(c)$ $dx$
$cint_b^adx=c[x]_b^a$
$=>c(a-b)$
$=>cxxa-bxxc="RHS"$

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