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I am told that the following equality follows from integration by parts:

$$\int_{R-k}^{1}(\theta-R)dG(\theta)-G(R-k)k=\int_{R-k}^{1}(1-G(\theta))d\theta-k$$ Where $R>k>0$ and $G$ is the CDF of $\theta$ which is distributed on $[0,1]$. Can someone explain how integration by parts has been used here? Thank you.

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Hint: Simply apply integration by parts to the integral on the LHS. Simplify and you should arrive at the following expression: \begin{equation} (1-R)-\int_{R-k}^1G(\theta)\mathrm d\theta. \end{equation} Add and subtract $k$ to obtain: \begin{equation} (1-R+k-k)-\int_{R-k}^1G(\theta)\mathrm d\theta = (1-(R-k))-k-\int_{R-k}^1G(\theta)\mathrm d\theta. \end{equation} Observe that $(1-(R-k))=\int_{R-k}^1(1)\mathrm d\theta$. Substitute that in and collect terms, you'll get the RHS expression.

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