integration by substitution

[mickeymouse]

If int { -1 to 3} f(x+k) dx= 8, where k is a constant, then int {(k-1) to (k+3)} f(x) dx= ???


I'm not sure how to approach this...

 

[GeneralSynopsis]

If int { -1 to 3} f(x+k) dx= 8, where k is a constant, then int {(k-1) to (k+3)} f(x) dx= ???


I'm not sure how to approach this...

Try \(\displaystyle u=x+k\)

.

 

[soroban]

Hello, mickeymouse!

If you can "read" what is given and make a sketch, the answer is clear.

\(\displaystyle \text{If }\:\int^3_{\text{-}1} f(x+k)\, dx\:= \:8\text{, where }k\text{ is a constant, find: }\:\int^{k+3}_{k-1} f(x)\, dx\)

\(\displaystyle \text{Consider the graph of }f(x)\text{ from }k-1\text{ to }k+3.\)

          |
          |     *         *
          |     :::*     *:
          |     ::::::*::::
          |     :::::::::::
      - - + - - * - - - - * - -
          |    k-1       k+3

\(\displaystyle \text{The graph of }f(x+k)\text{ is the same graph translated }k\text{ units to the left.}\)

          |
        * |       *
        ::|*     *:
        ::|:::*::::
        ::|::::::::
      - * + - - - * - - - -
       -1 |       3

\(\displaystyle \int^3_{\text{-}1}f(x+k)\,dx \:=\:8\:\text{ says that the area of the shaded region is 8 square units.}\)


\(\displaystyle \int^{k+3}_{k-1} f(x)\,dx\;\text{ asks for the area under the curve after it is moved }k\text{ units to the right.}\)
. . . . . . . . \(\displaystyle \text{as in the first diagram.}\)


\(\displaystyle \text{Obviously, the area is still 8 square units.}\)