Chapter 8: Q11-15P (page 459)

Use (11.6) and (11.14) to (11.16) to evaluate the following integrals.
(a) $\int_{0}^{π} s i n x δ (x - \frac{π}{2}) d x$
(b) $\int_{0}^{π} s i n x δ (x + \frac{π}{2}) d x$
(c) $\int_{- 1}^{1} e^{3 x} δ^{'} (x) d x$
(d) $\int_{0}^{π} c o s h x d^{''} (x - 1) d x$

Short Answer

Expert verified

The integration of given statements are

(a) $\int_{0}^{π} \sin x \cdot δ (x - \frac{π}{2}) d x$ is 1.

(b) $\int_{0}^{π} \sin x \cdot δ (x + \frac{π}{2}) d x$ is 0.

(d) $\int_{0}^{π} \cosh x \cdot d (x - 1) d x$ is cosh 1 .

Step by step solution

Given information

The given expression is

$(a) \int_{0}^{π} \sin x δ (x - \frac{π}{2}) d x (b) \int_{0}^{π} \sin x δ (x + \frac{π}{2}) d x (c) \int_{- 1}^{1} e^{3 x} δ^{'} (x) d x (d) \int_{0}^{π} \cosh x d^{''} (x - 1) d x$

Definition of Integration By Parts

Integration by partsor partial integration is a process that finds theintegralof aproductoffunctionsin terms of the integral of the product of theirderivativeandantiderivative.

(a) Calculate the Integration for ∫0πsinxδ(x-π2)dx

The calculation is

$\int_{0}^{π} \sin x δ (x - \frac{π}{2}) d x = (\sin x)_{s - x_{3}}$

Here $x_{0} = \frac{π}{2}, a = 0, b = π$ and $0 < \frac{π}{2} < π$

$= (\sin x)_{x - \frac{π}{2}} = \sin \frac{π}{2} = 1$

Thus, $\int_{0}^{π} \sin x \cdot δ (x - \frac{π}{2}) d x$ is 1.

(b) Calculate the Integration for ∫0πsinxδ(x+π2)dx

The calculation is

$\int_{0}^{π} \sin x δ (x + \frac{π}{2}) d x \int_{0}^{π} \sin x δ (x - (- \frac{π}{2})) d x$

Here $x_{0} = \frac{- π}{2}, a = 0, b = π$ and $\frac{- π}{2} \notin (0, π)$

$= 0$

Thus, $\int_{0}^{π} \sin x \cdot δ (x + \frac{π}{2}) d x$ is 0.

(c) Calculate the Integration for ∫-11e3xδ'(x)dx

The calculation is

$\int_{- 1}^{1} e^{3 x} δ' (x) d x = \int_{- 1}^{1} e^{3 x} \cdot δ' (x - 0) d x$

Where $ϕ (x) = e^{3x}$ and $a = 0$

$= - ϕ' (x) = - {(e^{3 x})}^{'} = - {[3 e^{3 x}]}_{x - 0} = - 3 e^{0} = - 3$

Thus, $\int_{- 1}^{1} e^{3 x} e^{'} (x) d x$ is -3.

(d) Calculate the Integration for ∫0πcoshxδ''(x-1)dx

The calculation is

$\int_{0}^{x} \cosh x \cdot δ'' (x - 1) d x$

Where $ϕ (x) = \cosh x$ and $a = 1$

$ϕ' (x) = (\cosh x)' = (\sinh x) ϕ'' (x) = (\sinh x)' = (\cosh x)_{x - 1} ϕ'' (x) = \cosh 1$

Thus, $\int_{0}^{x} \cosh x - δ^{''} (x - 1) d x$ is cosh1.

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Use (11.6) and (11.14) to (11.16) to evaluate the following integrals.
(a) $\int_{0}^{π} s i n x δ (x - \frac{π}{2}) d x$
(b) $\int_{0}^{π} s i n x δ (x + \frac{π}{2}) d x$
(c) $\int_{- 1}^{1} e^{3 x} δ^{'} (x) d x$
(d) $\int_{0}^{π} c o s h x d^{''} (x - 1) d x$

Short Answer

Step by step solution

Given information

Definition of Integration By Parts

(a) Calculate the Integration for ∫0πsinxδ(x-π2)dx

(b) Calculate the Integration for ∫0πsinxδ(x+π2)dx

(c) Calculate the Integration for ∫-11e3xδ'(x)dx

(d) Calculate the Integration for ∫0πcoshxδ''(x-1)dx

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Use (11.6) and (11.14) to (11.16) to evaluate the following integrals.(a)∫0πsinxδ(x-π2)dx(b)∫0πsinxδ(x+π2)dx(c)∫-11e3xδ'(x)dx(d)∫0πcoshxd''(x-1)dx

Short Answer

Step by step solution

Given information

Definition of Integration By Parts

(a) Calculate the Integration for ∫0πsinxδ(x-π2)dx

(b) Calculate the Integration for ∫0πsinxδ(x+π2)dx

(c) Calculate the Integration for ∫-11e3xδ'(x)dx

(d) Calculate the Integration for ∫0πcoshxδ''(x-1)dx

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Measurements

Atoms and Radioactivity

Oscillations

Thermodynamics

Electricity

Study anywhere. Anytime. Across all devices.

Use (11.6) and (11.14) to (11.16) to evaluate the following integrals.
(a) $\int_{0}^{π} s i n x δ (x - \frac{π}{2}) d x$
(b) $\int_{0}^{π} s i n x δ (x + \frac{π}{2}) d x$
(c) $\int_{- 1}^{1} e^{3 x} δ^{'} (x) d x$
(d) $\int_{0}^{π} c o s h x d^{''} (x - 1) d x$