Is this how a single integral works?

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Is this how a single integral works?

$\int_{b}^{a}$ $int_b^a$ $(c)$ $(c)$ $d x$ $dx$ $=$ $=$ $a$ $a$ $\times$ $xx$ c $-$ $-$ $b$ $b$ $\times$ $xx$ $c$ $c$

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Answer 1 · 2017-07-10T22:08:23

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

$\int_{b}^{a}$ $int_b^a$ $(c)$ $(c)$ $d x$ $dx$
$c \int_{b}^{a} d x = c {[x]}_{b}^{a}$ $cint_b^adx=c[x]_b^a$
$\Rightarrow c (a - b)$ $=>c(a-b)$
$\Rightarrow c \times a - b \times c = RHS$ $=>cxxa-bxxc="RHS"$

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Is this how a single integral works?

$\int_{b}^{a}$ $int_b^a$ $(c)$ $(c)$ $d x$ $dx$ $=$ $=$ $a$ $a$ $\times$ $xx$ c $-$ $-$ $b$ $b$ $\times$ $xx$ $c$ $c$

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Is this how a single integral works?

int_b^a∫abint_b^a(c)(c)(c)dxdxdx===aaaxx×xxc-−-bbbxx×xxccc

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$\int_{b}^{a}$ $int_b^a$ $(c)$ $(c)$ $d x$ $dx$ $=$ $=$ $a$ $a$ $\times$ $xx$ c $-$ $-$ $b$ $b$ $\times$ $xx$ $c$ $c$