Question

Let \[ f(t)= \begin{cases} 1, & t\in[0,2] \\ -t+3, & t\in[2,3] \\ 0, & \text{otherwise} \end{cases} \] Then \[ \int_{-\infty}^{\infty} f(\tau)\,d\tau = \_\_\_\_\_. \] (rounded off to one decimal place)

Hint:

Split the integral at the point where the piecewise definition changes, then add the areas of the two pieces.

Answer 1

Correct Answer: 2.5

Step 1: Understand what the integral means.
The integral of \(f(\tau)\) over all \(\tau\) is just the total area under the curve \(f(t)\), because \(f(t)\) is zero outside \([0,3]\) and only takes the two given pieces inside that range.

Step 2: Split the integral at the breakpoint.
Since \(f(t)\) is defined piecewise, split the range \([0,3]\) at \(t=2\):
\[ \int_{-\infty}^{\infty} f(\tau)\,d\tau = \int_0^2 1\,dt + \int_2^3 (-t+3)\,dt \]

Step 3: Evaluate the first piece.
This is the area of a rectangle of height 1 and width 2:
\[ \int_0^2 1\,dt = 2 \]

Step 4: Evaluate the second piece.
\[ \int_2^3 (-t+3)\,dt = \left[-\frac{t^2}{2}+3t\right]_2^3 \] At \(t=3\): \(-\frac{9}{2}+9=4.5\). At \(t=2\): \(-2+6=4\).
\[ \int_2^3 (-t+3)\,dt = 4.5-4 = 0.5 \] This piece is also just the area of a right triangle with base 1 and height 1, which gives \(0.5\) directly, matching the calculation.

Step 5: Add the two pieces.
\[ \int_{-\infty}^{\infty} f(\tau)\,d\tau = 2+0.5 = 2.5 \]

Final Answer:
The value of the integral is \[ \boxed{2.5} \]