Use the table of integrals to find the exact area of the region bounded by the graphs of the equ

step 1 Let R be the region bounded by the following curves. Y equal to x over square root of x plus 1,

Use the table of integrals to find the exact area of the...

Key Concepts

Definite Integration

Definite integration involves calculating the integral of a function between two specific limits, which gives the net area under the curve between those limits. This technique is essential for finding exact areas of regions bounded by curves in calculus.

Area Between Curves

The concept of finding the area between curves pertains to computing the region enclosed by one or more functions and the coordinate axes or other functions. This is achieved by setting up appropriate definite integrals that represent the vertical or horizontal slices of the region.

Integration Techniques

Integration techniques, such as substitution, are methods used to simplify complex integrals into forms that are easier to evaluate. Using a table of integrals often involves identifying parts of the integrand that match standard forms, thus facilitating the calculation of the area in problems where the integrand does not immediately fit basic integration rules.

Graphical Approximation

Graphical approximation involves using technology or graphing utilities to visually represent functions and estimate the area of a region. This approach serves to confirm the analytical results obtained through integration and to provide a numerical approximation when an exact expression is difficult to interpret.

Transcript

00:01 Let r be the region bounded by the following curves.

00:04 Y equal to x over square root of x plus 1, y equal 0, and x equal to 8.

00:10 Now the curves x over square root of x plus 1 and y equals 0, they intersect whenever x over the square root of x plus 1, this is equal to 0, or that means x is 0.

00:32 And so from here we can say that the region is in between 0 and 8.

00:39 Over here, we have the area of the region shaded in blue.

00:48 From here, we can see that the area of the region will be equal to the integral from 0 to 8 of x over the square root of x plus 1, which is the upper function, minus the bottom function, which is 0, dx.

01:09 And then this will simplify to the integral from 0 to 8 of x over the square root of x plus 1 dx.

01:19 Now we are ready to integrate.

01:23 Because the denominator has the form a plus b to the n, where n here is one -half, we then use formula 5 of appendix b, which is, comparing this formula and the integral representing a we have a equal to 1 b equal to 1 and n equal to 1 half and so using formula 5 we have 1 over 1 squared times negative 1 over n which is 1 half minus 2 times x plus 1 to the 1 half minus 2 plus 1 over 1 half minus 1 times x plus 1 raise to 1 half minus 1.

02:31 This evaluated from 0 to 8.

02:36 Now simplifying we have this part right here is just 1.

02:40 So we have negative 1 over negative 3 halves.

02:48 Times x plus 1 to the negative 3 halves plus 1 over that's negative 1 half times x plus 1 raised to negative 1 half evaluated from 0 to 8.

03:05 Simplifying further we get positive 2 over 3 times x plus 1 raised to positive 3 halves minus 2 times x plus 1 raise to positive 1 half evaluated from 0 to 8 now when x is 8 we have 2 over 3 times 8 plus 1 raise to 3 halves minus 2 times 8 plus 1 raised to 1 half and when x is 0 we have 2 over 3 times 0 plus 1 to the 3 halves minus 2 times 0 plus 1 raise to 1 half.

03:49 This then gives us 2 over 3 times 9 to the 3 halves minus 2 times 9 to the 1 half minus 2 over 3 times 1 to the 3 halves plus 2 times 1 to the 1 half...

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