Calculus - definite integral

In summary, the conversation is about solving the definite integral ∫202x(4−x2)1/5 dx and determining the most efficient substitution to use. The suggested substitution is u=4-x^2 and with that, the integral becomes of the form ∫baf(u)du with a=4 and b=0. The integrand is 2x(4-x^2)^{1/5}dx which can be rewritten as -(4-x^2)^{1/5}(-2xdx). Using the substitution, the integral becomes -∫u^{1/5}du. Finally, the exact value of ∫202x(4−x2)1∕5 dx
  • #1
Yousra1
1
0
Consider the definite integral ∫202x(4−x2)1/5 dx.
What is the substitution to use? u= 4-x^2
Preview Change entry mode (There can be more than one valid substitution; give the one that is the most efficient.)
For this correct choice, du/dx= -2x
Preview Change entry mode
If we make this substitution, then the integral becomes of the form ∫baf(u)du. What are a, b and f(u)?
a= 4
b= 0
f(u)=
Preview Change entry mode
Finally, use this work to compute

∫202x(4−x2)1∕5 dx=
Preview Change entry mode
Give the exact value.
 
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  • #2
Re: calculus

What is this? Did you just post what appeared on a computer screen?

[tex]
Yousra said:
Consider the definite integral ∫202x(4−x2)1/5 dx.
What you have written is [tex]\int 202 x(4- 2x)(1/5)dx[/tex]. That's not a "definite integral" because there are no limits of integration.

But from what there is below it appears you mean [tex]\int_2^0 2x(4- x^2)^{1/5}dx[/tex]. Is that right?

If you cannot use Latex, at least use the standard ASCII "_" for subscripts and "^" for superscripts:
integral_2^0 2x(4- x^2)^(1/5) dx

What is the substitution to use? u= 4-x^2
Preview Change entry mode (There can be more than one valid substitution; give the one that is the most efficient.)
For this correct choice, du/dx= -2x
Preview Change entry mode
What in the world is "Preview Change entry mode"? Is this just copied arbitrarily from the computer screen?
Yes, if you make the substitution u= 4- x^2 them du= -2x dx

If we make this substitution, then the integral becomes of the form ∫baf(u)du. What are a, b and f(u)?
a= 4
b= 0
f(u)=
Yes, when x= 0, u= 4- 0^2= 4 and when x= 2, u= 4- 2^2= 0. The integrand appears to be [tex]2x(4- x^2)^{1/5}dx[/tex] which can be written [tex](4- x^2)^{1/5}(2xdx)= -(4- x^2)^{1/5}(-2xdx)[/tex].
Since [tex]4- x^2= u[/tex] and [tex]-2xdx= du[/tex], that is [tex]-u^{1/5}du[/tex].

Preview Change entry mode
Finally, use this work to compute

∫202x(4−x2)1∕5 dx=
Preview Change entry mode
Give the exact value.
 

Related to Calculus - definite integral

1. What is a definite integral?

A definite integral is a mathematical concept in calculus that represents the area under a curve between two specified points on the x-axis. It is denoted by ∫ab f(x) dx, where a and b are the limits of integration, f(x) is the integrand, and dx represents the infinitesimal change in x.

2. How is a definite integral different from an indefinite integral?

While a definite integral has specific limits of integration and gives a numerical value, an indefinite integral does not have limits and gives a function as the result. In other words, a definite integral calculates the area under a curve, while an indefinite integral finds the original function.

3. What is the fundamental theorem of calculus?

The fundamental theorem of calculus states that the definite integral of a function f(x) can be calculated by finding the antiderivative of f(x) and evaluating it at the limits of integration. In other words, integration and differentiation are inverse processes.

4. What are some real-world applications of definite integrals?

Definite integrals have various applications in physics, engineering, and economics. For example, they can be used to calculate the work done by a force, the displacement of an object, and the total cost of production. They are also used in statistics to calculate probabilities and in finance to determine compound interest.

5. How can the definite integral be interpreted geometrically?

Geometrically, the definite integral represents the area of a region bounded by the x-axis, the curve, and the vertical lines at the limits of integration. It can also be interpreted as the net signed area, where areas above the x-axis are positive and areas below the x-axis are negative.

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