The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral

The graph of g consists of two straight lines and a...

Key Concepts

Definite Integral

The definite integral of a function over an interval represents the net accumulation of quantities, which can be interpreted as the signed area between the function's graph and the x-axis. It is foundational in understanding how integration sums up small contributions across an interval.

Geometric Interpretation of Integration

When a function's graph is composed of simple geometric shapes such as lines and curves, the integral can be evaluated by calculating the area of these shapes directly. This approach simplifies the process by using familiar formulas from geometry to determine the area under the curve.

Piecewise Functions and Decomposition

Many functions are defined in a piecewise manner, meaning they have different expressions on different intervals. In such cases, the integral over a larger interval is computed by breaking it down into a sum of integrals over smaller intervals where the function's behavior is uniform or follows a simple pattern.

Area Formulas for Basic Shapes

Knowing the formulas for areas of basic geometric shapes such as triangles, rectangles, circles, and semicircles is essential when using geometric methods to evaluate integrals. These formulas allow for quick calculation of areas corresponding to the regions under the curve, thereby facilitating the evaluation of the overall integral.

Transcript

00:01 The graph of a function g consists of two straight lines and a semicircle, as we see in the following graph.

00:11 With that information, we want to evaluate the integrals in part a, the interval from 0 to 2 of g of x, differential of x, in part p, the integral of g between 2 and 6, and in part c, the integral of g between 0 and 7.

00:32 So we're going to use the information about what kind of graph we have in each part to calculate the integrals in a very straightforward way.

00:46 So in part a, we are going to talk about the integral from 0 to 2 of g of x.

00:57 And from 0 to 2, we have the first straight line.

01:02 We have the other straight line on the interval 06 -7.

01:07 The semicircle is drawn between 2 and 6 as we see here.

01:13 And this thread line here on the interval 0 ,2, passes through the point 04 here and through the point to 0.

01:33 So we know that this interval here represents the area under the graph that is under the straight line and above the axis.

01:56 Within or over if you want the interval 02 so this integral corresponds to the area of a right triangle is exactly that to the area of the right triangle with vertices 0 here to 0 here and 04 here is the right triangle and the area we know how to calculate it directly the area is one half times the height of the triangle that is four units as we see here one two three four units times the base of the triangle which is two units we cancel out the twos and we get four so the integral from 0 to 2 of g is 4 because we saw that the area corresponds to the area of the right triangle which can be calculated directly.

04:01 Now in part b we are going to calculate the interval from 2 to 6 of g.

04:08 And over this interval the function is a semicircle, that is a semicircle, that is completely under the x -axis.

04:18 So the semicircle has a center, this point here, this is the center of the circle, and the center then is for zero.

04:48 And the radius of the circle is 2.

04:54 To unit and radius 2.

05:04 So we can write the equation in the following way.

05:11 So any point xy is on the circle if and only if the square root of that is the distance of the point xy to the point four zero there is the square root of x minus four square plus y minus zero square is equal to the radius that is two this equation is equivalent to x minus four square plus y square equals four which is equivalent to four square equals four minus four minus x minus 4 square this is 4 minus x square minus 8 x plus 16 that is 4 minus x square plus 8 x minus 16 and that is negative x square plus 8 x minus 12 and that is y square so it means that we if we consider the complete circle we don't have because any point could have two images that's not a function but if we take only a semicircle like this like in this case we will have a function that is the portion of the function or the piece of the function that is over the interval to six only the semicircle which is below the x -axis and because it's below the x -axis is always negative so we know that that corresponds to the value of y, which we take from this equation, by taking the negative square root of this expression.

07:20 So, so the semi -circle used in g is y equals negative square root of negative, square root of x, negative x, negative x squared, sorry, plus x, 8x, minus 12.

07:45 This is the equation because is the result of solving for y the equation of the equation of the circle and considering that the square root is got to be taken with negative sign because the graph of the semicircle is completely below the y -axis and so this will be the equation but we don't have to make the calculations that is the integral from 2 to 6 of g of x is true that is equal to negative integral from 2 to 6 of the square root of negative x square plus 8x minus 12 and this will be a nightmare to solve this interval here but it is not necessary because we use the following reasoning we know that the area of or enclosed by the graph of the function over 2 6 that is the semicircle the x -axis and within 2 -6 is the half of the area of the circle but with negative sign because the integral is going to be negative because the function is negative so we can say here this way since g of x is less than or equal to zero for x over the interval to six then we know that the integral from 2 to 6 of g is a negative number or 0 and we know that the magnitude of this integral if we take the absolute value that is half of the area of the entire circle as we saw here we put the positive part of the circle we will get the area of the whole circle but in this case we are going to have the area of half the circle but the integral itself is going to be negative because the function is negative over all the interval to six that's very important to notice because you can think of the area but it's important to notice what sign will be will have the integral when we solve directly the intro and not thinking about the area only.

10:52 So in this case we know that the area of the circle is pi radius square.

11:05 That is pi times the radius of a circle we know is 2 we saw here on the graph so it's 2 square and that is 4 pi.

11:19 Then the half half of the the area of the circle which is in fact the area of a semi -circle is 4 pi over 2 that is 2 pi then considering that the integral is negative number the integral from 2 to 6 of g is equal to negative to pi that is negative to pi that is is negative, but the magnitude and the number is half of the area of the circle, which is 2 pi.

12:10 So we have already calculated this part b.

12:14 Now we go to bar c where we want to calculate the interval from 0 to 7 of g of xx.

12:22 And that we know is equal to the integral.

12:25 We saw the graph again...

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