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

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

Key Concepts

Area Under a Curve

Calculating the area under a curve involves integrating the function that represents the boundary of the region with respect to the independent variable. In the context of finding enclosed areas, the definite integral provides an exact measure of the area between the curve and a baseline, typically the x-axis.

Standard Integral Formulas

Using standard integral formulas, often found in tables of integrals, allows one to solve integrals that might otherwise be challenging. This technique involves matching the integrand to a known form and applying the corresponding antiderivative directly, simplifying the process of finding an exact answer.

Definite Integration

Definite integration is the process of computing the integral of a function over a specific interval, which yields the net area between the function's graph and the x-axis. This concept is fundamental in calculus for calculating areas of regions bounded by curves and lines over a closed interval.

Graphical Estimation

Graphical estimation involves using graphing utilities to visualize the region described by the functions and to approximate the area related to it. This approach serves as a valuable tool for verifying analytical results obtained through integration by providing a visual confirmation of the area calculation.

Transcript

00:01 Let r be the region bounded by y equals x squared times the square root of x squared plus 4, y equals 0, and x equals the square root of 5.

00:10 Now, these two curves will only intersect whenever they are equal.

00:16 That means when x squared times the square root of x squared plus 4 equal 0.

00:24 Now, this implies that x squared is 0 or that the square root of x squared plus 4 equals 0.

00:31 Plus 4 is 0.

00:33 Now x squared equals 0, this implies that x is 0, and the square rate of x4 equals 0, this implies that x squared plus 4 is 0.

00:44 Now, this further implies that x squared is equal to negative 4, which is not true because x squared is always positive.

00:55 And so, the only intersection is at x equals 0.

01:00 That means the region is found in between x is 0 and x equals the square rate of 5.

01:07 Below is the graph of the region.

01:12 In this graph we will see that the area is equal to the integral from 0 to square root of 5 of the top function x squared times the square out of x squared plus 4 minus the bottom function is 0.

01:28 Now this simplifies to the integral from 0 to square root of 5 of x squared times the square of x4 dx.

01:39 Now this integral follows the formula 27 of appendix b, which is written below.

01:47 From here we can say that x squared is equal to g squared or that means x is u.

01:56 And our a squared, this is equal to 4.

02:02 And we need a to the 4th, so this 8 to the 4th is also the square of 4, which is 16.

02:11 And so from here we have 1 over 8 times x times 2x squared plus 4 times the square of x squared plus 4 minus 16 natural log.

02:32 Of the absolute value of x plus the square root of x squared plus four, evaluated from 0 to square root of 5.

02:42 Now when next is square root of 5, we have 1 over 8 times square root of 5 times 2 times square of 5 squared plus 4 times the square of square root of 5 squared plus 4 minus 16 natural log of the absolute value of square root of 5 plus the square root of 5 plus 4 and then we subtract this when x is 0 we have 1 over 8 times 0 times 2 0 2 0 squared 2 plus 4 times the square of 0 squared plus 4 minus 16 natural log of the absolute value of 0 plus the square root of 0 plus 4.

03:42 Simplifying we get 1 over 8 times square root of 5 times 14 times the square root of 9 minus 16.

03:55 Natural log of the absolute value of square root of 5 plus square root of 9.

04:03 And then subtract this by 1 over 8 times that's 0 minus 16 natural log of the square of 4, which is 2.

04:17 This then gives us 1 over 8 times square root of 5 times 14 times 3 minus 16 natural log of the absolute value of square root of five plus three minus one over eight times negative 16 natural log of two.

04:42 This will then be equal to one over eight times 42 square root of five minus 16 natural log of the absolute value of square root of five plus three minus or plus 1 over 8 times 16 natural log of 2...

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