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The given integral is${鈭�}_{-鈭�}^{C}f\left(x\right)dx+{鈭�}_c^{鈭�}f\left(x\right)dx.$

To prove that for all real numbers d,${鈭�}_{-鈭�}^{C}f\left(x\right)dx+{鈭�}_c^{鈭�}f\left(x\right)dx$is equal to ${鈭�}_{-鈭�}^{d}f\left(x\right)dx+{鈭�}_d^{鈭�}f\left(x\right)dx.$ Let the function f is continuous on [a,b] and for any real number c,${鈭�}_a^{b}f\left(x\right)dx={鈭�}_a^{C}f\left(x\right)dx+{鈭�}_c^{b}f\left(x\right)dx.$

We will prove the given statements in two cases when c < d and whend < c.

We will use the property of definite integrals,

$$\begin{gathered} {鈭�}_{-鈭�}^{C}f\left(x\right)dx+{鈭�}_c^{鈭�}f\left(x\right)dx \\ ={鈭�}_{-鈭�}^{C}f\left(x\right)dx+{鈭�}_c^{d}f\left(x\right)dx+{鈭�}_d^{鈭�}f\left(x\right)dx \\ \mathrm{Use}{鈭�}_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={鈭�}_{\mathrm{a}}^{\mathrm{C}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+{鈭�}_{\mathrm{c}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ ={鈭�}_{-鈭�}^{\mathrm{C}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+{鈭�}_{\mathrm{c}}^{鈭�}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \end{gathered}$$

We will use the property of definite integrals,

$$\begin{gathered} {鈭�}_{-鈭�}^{d}f\left(x\right)dx+{鈭�}_d^{鈭�}f\left(x\right)dx \\ ={鈭�}_{-鈭�}^{d}f\left(x\right)dx+{鈭�}_d^{c}f\left(x\right)dx+{鈭�}_c^{鈭�}f\left(x\right)dx \\ \mathrm{Use}{鈭�}_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={鈭�}_{\mathrm{a}}^c\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+{鈭�}_{\mathrm{c}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ ={鈭�}_{-鈭�}^{\mathrm{C}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+{鈭�}_{\mathrm{c}}^{鈭�}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \end{gathered}$$

Thus, role="math" localid="1649143526896" ${鈭�}_{-鈭�}^{c}f\left(x\right)dx+{鈭�}_c^{鈭�}f\left(x\right)dx={鈭�}_{-鈭�}^{d}f\left(x\right)dx+{鈭�}_d^{鈭�}f\left(x\right)dx.$

Hence proved.