Definite Integral of Function plus Constant

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Theorem

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.

Let $c$ be a constant.


Then:

$\ds \int_a^b \paren {\map f t + c} \rd t = \int_a^b \map f t \rd t + c \paren {b - a}$


Proof

Let $P = \set {x_0, x_1, x_2, \ldots, x_n}$ be a finite subdivision of $\closedint a b$.

Let $\map {L^{\paren {f + c} } } P$ be the lower Darboux sum of $\map f x + c$ on $\closedint a b$ belonging to $P$.

Let:

$\ds m_k^{\paren {f + c} } = \map {\inf_{x \mathop \in \closedint {x_{k - 1} } {x_k} } } {\map f x + c}$

where $k \in \set {0, 1, \ldots, n}$.


So:

\(\ds m_k^{\paren {f + c} }\) \(=\) \(\ds \map {\inf_{x \mathop \in \closedint {x_{k - 1} } {x_k} } } {\map f x + c}\)
\(\ds \) \(=\) \(\ds c + \map {\inf_{x \mathop \in \closedint {x_{k - 1} } {x_k} } } {\map f x}\)
\(\ds \) \(=\) \(\ds c + m_k^{\paren f}\)


It follows that:

\(\ds \map {L^{\paren {f + c} } } P\) \(=\) \(\ds \sum_{k \mathop = 1}^n {m_k^{\paren {f + c} } } \paren {x_k - x_{k - 1} }\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n m_k^{\paren f} \paren {x_k - x_{k - 1} } + c \sum_{k \mathop = 1}^n \paren {x_k - x_{k - 1} }\)
\(\ds \) \(=\) \(\ds \map {L^{\paren f} } P + c \paren {b - a}\) as $\ds \sum_{k \mathop = 1}^n \paren {x_k - x_{k - 1} }$ telescopes


So from the definition of definite integral, it follows that:

\(\ds \int_a^b \paren {\map f t + c} \rd t\) \(=\) \(\ds \map {\sup_P} {\map {L^{\paren {f + c} } } P}\)
\(\ds \) \(=\) \(\ds \map {\sup_P} {\map {L^{\paren f} } P + c \paren {b - a} }\)
\(\ds \) \(=\) \(\ds \map {\sup_P} {\map {L^{\paren f} } P} + c \paren {b - a}\)
\(\ds \) \(=\) \(\ds \int_a^b \map f t \rd t + c \paren {b - a}\)

$\blacksquare$


Sources

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