Integral Calculation Strategy

To find the definite integral $\int_{-\infty}^{\infty} f(\tau) d\tau$, we need to consider the intervals where the function $f(t)$ is non-zero.

The function $f(t)$ is defined as:

Since $f(t) = 0$ outside the interval $[0, 3]$, the integral simplifies to:

We can split this integral into two parts based on the definition of $f(t)$.

Step-by-Step Integral Evaluation

• Step 1: Integrate over $[0, 2]$

  In this interval, $f(\tau) = 1$.

  $ \int_{0}^{2} f(\tau) d\tau = \int_{0}^{2} 1 d\tau $

  Evaluating the integral:

  $ [\tau]_{0}^{2} = 2 - 0 = 2 $
• Step 2: Integrate over $[2, 3]$

  In this interval, $f(\tau) = -\tau + 3$.

  $ \int_{2}^{3} f(\tau) d\tau = \int_{2}^{3} (-\tau + 3) d\tau $

  Evaluating the integral:

  $ \left[-\frac{\tau^2}{2} + 3\tau\right]_{2}^{3} = \left(-\frac{3^2}{2} + 3(3)\right) - \left(-\frac{2^2}{2} + 3(2)\right) $ $ = \left(-\frac{9}{2} + 9\right) - \left(-2 + 6\right) $ $ = \left(\frac{9}{2}\right) - (4) = 4.5 - 4 = 0.5 $
• Step 3: Sum the results

  Add the results from Step 1 and Step 2 to get the total integral value.

  $ \int_{-\infty}^{\infty} f(\tau) d\tau = \int_{0}^{2} f(\tau) d\tau + \int_{2}^{3} f(\tau) d\tau $ $ = 2 + 0.5 = 2.5 $
• Step 4: Round the result

  The question asks for the result rounded to one decimal place. The calculated value is $2.5$, which is already in the required format.

  The value $2.5$ lies between 2.4 and 2.6.