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)
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:
$f(t) = \begin{cases} 1, & t \in [0, 2] \\ -t + 3, & t \in [2, 3] \\ 0, & \text{otherwise} \end{cases}$Since $f(t) = 0$ outside the interval $[0, 3]$, the integral simplifies to:
$\int_{-\infty}^{\infty} f(\tau) d\tau = \int_{0}^{3} f(\tau) d\tau$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.
Important Questions from Definite Integrals
The Legendre polynomials P n(x), n = 0, 1, 2, ..., satisfying the orthogonailty condition \(\int_{{\rm{ - 1}}}^{\rm{1}} {{{\rm{P}}_{\rm{n}}}\left( {\rm{x}} \right){{\rm{P}}_{\rm{m}}}} \left( {\rm{x}} \right){\rm{dx}}\,{\rm{ = }}\,\frac{{\rm{2}}}{{{\rm{2n + 1}}}}{{\rm{\delta }}_{{\rm{nm}}}}\) on the interval [-1, +1], may be defined by the Rodrigues formula P n(x) = \(\frac{{\rm{1}}}{{{{\rm{2}}^{\rm{n}}}{\rm{n!}}}}\frac{{{{\rm{d}}^{\rm{n}}}}}{{{\rm{d}}{{\rm{x}}^{\rm{n}}}}}{\left( {{{\rm{x}}^{\rm{2}}}{\rm{ - 1}}} \right)^{\rm{n}}}\) . The value of the definite integral \(\int_{{\rm{ - 1}}}^{\rm{1}} {\left( {{\rm{4 + 2x - 3}}{{\rm{x}}^{\rm{2}}}{\rm{ + 4}}{{\rm{x}}^{\rm{3}}}} \right){{\rm{P}}_{\rm{3}}}\left( {\rm{x}} \right){\rm{dx}}} \) is
A parametric curve is defined \(x = cos\left(\frac{\Pi t}{2}\right) , Y= sin\left(\frac{\Pi t}{2}\right)\) in the range of \(0\leq t\leq 1\) . It is rotated about X-axis by 360°.
‘What is the area of the surface generated?
if \(\displaystyle\int\dfrac{\sin x}{\sin (x-a)}dx=Ax+B\log |sin(x-a)|+ C\) where A, B and c are real constants then:
Which of the following is NOT a property of definite integral?
The value of integral \(x\over cos^2 x\) is equal to
