Question

It \(f\)is continuous on, prove that

\(\int\limits_a^b {f(x + c)dx = } \int\limits_{a + c}^{b + c} {f(x)dx} \)

For the case where \(f(x) \ge 0\) draw a diagram to interpret this equation geometrically as equality of area.

Short answer

Hence proved \(\int\limits_a^b {f(x + c)dx = } \int\limits_{a + c}^{b + c} {f(x)dx} \)

Step-by-step solution

Solution of LHS of the equation

\(\begin{aligned}{l}\int\limits_a^b {f(x + c)dx &= } f\int\limits_a^b {c\;dx\; + \;x\;dx} \\ &= f(\int\limits_a^b {c\;dx\; + \;\int\limits_a^b {x\;dx} } )\end{aligned}\)

\( = f\left( {cb - ca + \frac{{{b^2} - {a^2}}}{2}} \right)\)

Solution of RHS of the equation

\(\begin{aligned}{l}\int\limits_{a + c}^{b + c} {f(x)dx &= } f\int\limits_{a + c}^{b + c} {x\;dx} \\ &= f\left( {\frac{{{x^2}}}{2}} \right)_{a + c}^{b + c}\end{aligned}\)

\( = f\frac{2}{2}\left( {cb - ca + \frac{{{b^2} - {a^2}}}{2}} \right) = f\left( {cb - ca + \frac{{{b^2} - {a^2}}}{2}} \right)\)

Step 3: Final Proof

From Step 1 and Step 2 we get that answers for both as same that is

\(\int\limits_a^b {f(x + c)dx = } \int\limits_{a + c}^{b + c} {f(x)dx} = f\left( {cb - ca + \frac{{{b^2} - {a^2}}}{2}} \right)\)

\(\int\limits_a^b {f(x + c)dx = } \int\limits_{a + c}^{b + c} {f(x)dx} \)