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Question
`int_("a" + "c")^("b" + "c") "f"(x) "d"x` is equal to ______.
Options
`int_"a"^"b" "f"(x - "c")"d"x`
`int_"a"^"b" "f"(x + "c")"d"x`
`int_"a"^"b" "f"(x)"d"x`
`int_("a" - "c")^("b" - "c") "f"(x)"d"x`
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Solution
`int_("a" + "c")^("b" + "c") "f"(x) "d"x` is equal to `int_"a"^"b" "f"(x + "c")"d"x`.
Explanation:
Since by putting x = t + c, we get
I = `int_"a"^"b" "f"("c" + "t")"dt"`
= `int_"a"^"b" "f"(x + "c")"d"x`.
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