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Question
Prove that: `int_"a"^"b" "f"(x) "d"x = int_"a"^"c""f"(x) "d"x + int_"c"^"b" "f"(x) "d"x`, where a < c < b
Solution
Let `int "f"(x) "d"x = "g"(x) + c`
`int_"a"^"b""f"(x)"d"x = ["g"(x) + "c"]_"a"^"b"`
= `[{"g"("b") + "c"} - {"g"("a") + "c"}]`
= `"g"("b") - "g"("a")` ........(i)
`int_"a"^"c" "f"(x)"d"x + int_"c"^"b" "f"(x)"d"x = ["g"(x) + "c"]_"a"^"c" + ["g"(x) + "c"]_"c"^"b"`
= `[{"g"("c") + "c"} - {"g"("a") + "c"}] + ["g"("b") + "c"} - "g"("c") + "c"]_"c"^"b"`
= `"g"("c") + "c" - "g"("a") - "c" + "g"("b") + "c" - "g"("c") - "c"`
= `"g"("b") - "g"("a")` .......(ii)
From (i) and (ii), we get
`int_"a"^"b" "f"(x) "d"x = int_"a"^"c""f"(x) "d"x + int_"c"^"b" "f"(x) "d"x`, where a < c < b
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