Chapter 4: Problem 2

Consider the function $f(t)=2 t^{4}$. Evaluate the following integrals: (a) $\int_{0}^{10} f(t) \mathcal{H}(t-3) d t$ (b) $\int_{0}^{10} f(t) \dot{\mathcal{H}}(t-3) d t$ (c) $\int_{0}^{10} f(t) \ddot{\mathcal{H}}(t-3) d t$ (d) $\int_{0}^{10} \ddot{f}(t) \mathcal{H}(t-3) d t$

Short Answer

Expert verified

a) Integral with Heaviside function: $$\int_{0}^{10} f(t) \mathcal{H}(t-3) d t = \boxed{39902.8}$$ b) Integral with first derivative of Heaviside function: $$\int_{0}^{10} f(t) \dot{\mathcal{H}}(t-3) d t = \boxed{162}$$ c) Integral with second derivative of Heaviside function: $$\int_{0}^{10} f(t) \ddot{\mathcal{H}}(t-3) d t = \boxed{0}$$ d) Integral of second derivative of f(t) multiplied by Heaviside function: $$\int_{0}^{10} \ddot{f}(t) \mathcal{H}(t-3) d t = \boxed{7784}$$

Step by step solution

Use properties of the Heaviside function

When multiplying f(t) by $\mathcal{H}(t-3)$, the Heaviside function reduces the function to zero for all values of $t<3$. Therefore, the integral can be adjusted as follows: $$\int_{0}^{10} f(t) \mathcal{H}(t-3) d t = \int_{3}^{10} f(t) d t$$

Evaluate the integral

Now, evaluate the integral of the given function: $$\int_{3}^{10} 2t^4 dt = \frac{2}{5}t^5\Big|_3^{10} = \frac{2}{5}(10^5 - 3^5) = \frac{2}{5}(100000 - 243) = \frac{2}{5}(99757) = 39902.8$$ The value of the integral $\int_{0}^{10} f(t) \mathcal{H}(t-3) d t$ is 39902.8. #b) Integral with first derivative of Heaviside function#

Use properties of the Dirac delta function

Because the Dirac delta function is nonzero only at $t=3$, we can evaluate the integral as follows: $$\int_{0}^{10} f(t) \delta(t-3) dt = f(3) = 2 \cdot 3^4 = 2 \cdot 81 = 162$$ The value of the integral $\int_{0}^{10} f(t) \dot{\mathcal{H}}(t-3) d t$ is 162. #c) Integral with second derivative of Heaviside function#

Evaluate the integral with $\dot{\delta}$

The integral of delta function derivative is zero when the function is continuous at the point $t=3$. In this case, $f(t)$ is continuous, so the answer is: $$\int_{0}^{10} f(t) \ddot{\mathcal{H}}(t-3) dt = 0$$ The value of the integral $\int_{0}^{10} f(t) \ddot{\mathcal{H}}(t-3) d t$ is 0. #d) Integral of second derivative of f(t) multiplied by Heaviside function#

Find the second derivative of f(t)

First, find the derivative of f(t): $$f'(t) = \frac{d}{dt}(2t^4) = 8t^3$$ Now, find the second derivative: $$f''(t) = \frac{d}{dt}(8t^3) = 24t^2$$

Apply Heaviside function properties to integral

Use the same logic as in part a): $$\int_{0}^{10} f''(t) \mathcal{H}(t-3) dt = \int_{3}^{10} 24t^2 dt$$

Evaluate the integral

Now, evaluate the integral: $$\int_{3}^{10} 24t^2 dt = 8t^3\Big|_3^{10} = 8(10^3 - 3^3) = 8(1000 - 27) = 8\cdot973 = 7784$$ The value of the integral $\int_{0}^{10} \ddot{f}(t) \mathcal{H}(t-3) d t$ is 7784. Summarizing the results: (a) $\int_{0}^{10} f(t) \mathcal{H}(t-3) d t = 39902.8$ (b) $\int_{0}^{10} f(t) \dot{\mathcal{H}}(t-3) d t = 162$ (c) $\int_{0}^{10} f(t) \ddot{\mathcal{H}}(t-3) d t = 0$ (d) $\int_{0}^{10} \ddot{f}(t) \mathcal{H}(t-3) d t = 7784$

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Consider the function \(f(t)=2 t^{4}\). Evaluate the following integrals: (a) \(\int_{0}^{10} f(t) \mathcal{H}(t-3) d t\) (b) \(\int_{0}^{10} f(t) \dot{\mathcal{H}}(t-3) d t\) (c) \(\int_{0}^{10} f(t) \ddot{\mathcal{H}}(t-3) d t\) (d) \(\int_{0}^{10} \ddot{f}(t) \mathcal{H}(t-3) d t\)

Short Answer

Step by step solution

Use properties of the Heaviside function

Evaluate the integral

Use properties of the Dirac delta function

Evaluate the integral with \(\dot{\delta}\)

Find the second derivative of f(t)

Apply Heaviside function properties to integral

Evaluate the integral

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