Unique's questions at Yahoo Answers regarding integration

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In summary, we are asked to find the integrals of four different functions. The first three integrals are evaluated using different rules and properties, while the last one requires a substitution. The final values for the integrals are 2, -1/(4(x^4+2)), 0, and 1/2 respectively.
  • #1
MarkFL
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Here are the questions:

Find the following integrals please show steps?

find the following integrals

a)integral -pi/2,pi/2(2t+cost)dt

b)integral x^3/(x^4+2)^2 dx

c)integral -2,2x(x^2+3)^2 dx

d)integral 0,pi/4 tan x sec^2 x dx

I have posted a link there to this topic so the OP can see my work.
 
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  • #2
Hello unique,

We are given to evaluate:

a) \(\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}2t+\cos(t)\,dt\)

I would write this as the sum of two integrals:

\(\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}2t\,dt+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(t)\,dt\)

Now, by the odd and even rules, we may write:

\(\displaystyle I=0+2\int_{0}^{\frac{\pi}{2}}\cos(t)\,dt=2\left[\sin(t) \right]_{0}^{\frac{\pi}{2}}=2(1-0)=2\)

b) \(\displaystyle I=\int\frac{x^3}{\left(x^4+2 \right)^2}\,dx\)

Let:

\(\displaystyle u=x^4+2\,\therefore\,du=4x^3\,dx\)

and we have:

\(\displaystyle I=\frac{1}{4}\int u^{-2}\,du=-\frac{1}{4u}+C=-\frac{1}{4\left(x^4+2 \right)}+C\)

c) \(\displaystyle I=\int_{-2}^2 x(x^2+3)^2\,dx\)

By the odd-function rule, we have:

\(\displaystyle I=0\)

d) \(\displaystyle I=\int_0^{\frac{\pi}{4}} \tan(x)\sec^2(x)\,dx\)

Let:

\(\displaystyle u=\tan(x)\,\therefore\,du=\sec^2(x)\,dx\)

and we have:

\(\displaystyle I=\int_0^1 u\,du=\frac{1}{2}\left[u^2 \right]_0^1=\frac{1}{2}(1-0)=\frac{1}{2}\)
 

FAQ: Unique's questions at Yahoo Answers regarding integration

1. What is integration in mathematics?

Integration is a mathematical concept used to find the area under a curve or the accumulation of a quantity over a given interval. It is the inverse operation of differentiation and is typically used in calculus and other advanced math courses.

2. How is integration used in real life?

Integration has many practical applications in fields such as physics, engineering, and economics. For example, it is used to calculate the volume of irregularly shaped objects, find the center of mass of an object, and determine the total revenue or profit of a business over a certain time period.

3. What are the different methods of integration?

The most common methods of integration include substitution, integration by parts, partial fractions, trigonometric substitution, and numerical integration. Each method is used to solve different types of integrals and can be chosen based on the complexity of the function being integrated.

4. Why is integration important?

Integration is a fundamental concept in mathematics and is essential for solving many real-world problems. It allows us to find precise solutions to complex mathematical equations and provides a way to model and understand the behavior of various systems and processes.

5. Can integration be used in multiple dimensions?

Yes, integration can be used in multiple dimensions. In addition to one-dimensional integration, which is commonly used to find the area under a curve, there is also double and triple integration, which are used to find the volume under a surface and the volume of a solid, respectively. These higher dimensional integrals are often used in physics and engineering applications.

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