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find the area of the region bounded by the graph of the function

0 votes
1. (a) Find the area of the region bounded by the graph of the function f(x) = x^2 + 3, the X-axis and the lines x = −1 to x = 2.

    (b) Find the area of the region bounded by the graph of the function f(x) = e^x + 2, the X-axis and the lines x = 0 to x = 3.

 

 2. (a) Find the area of the region enclosed by functions f(x) = x^2 and g(x) = x

     (b) Find the area of the region bounded by the graph of the function f(x) = x^2 − 9 and g(x) = 9 − x^2 .

 

3. Find the area of the region bounded by the graph of the function f(x) = x^2−4, the X-axis and the lines x = −4 to x = 3.
asked Oct 4, 2015 in CALCULUS by anonymous

5 Answers

0 votes

1(a)

Step 1:

The curves are image and image-axis and the lines are image, image.

Area of the region between two curves:

If image and image are continuous on image and image for all image in image, then the area of the region bounded by the graphs of image and image and the vertical lines image and image is image.

Here, image, image, image and image.

Graph:

Graph the functions image, image and the vertical lines image and image.

image

Observe the graph:

The two functions on the interval image.

Step 2:

Find the area of the region.

Substitute image, image and image in image.

image

image

Apply power rule of integration: image.

image

image

image

image

image

image

image.

Area bounded by the two curves is image.

Solution:

Area bounded by the two curves is image.

answered Oct 5, 2015 by Sammi Mentor
0 votes

1(b)

Step 1:

The curves are image, image-axis and the lines are image, image.

The area of the region bounded by the graphs of image and image and the vertical lines image and image is image.

Here, image, image, image and image.

Graph:

Graph the functions image, image and the vertical lines image and image.

image

Observe the graph:

The two functions on the interval image.

Step 2:

Find the area of the region.

Substitute image, image and image in image.

image

image

Basic integral formula: image.

image

image

image

image

image.

Area bounded by the two curves is image.

Solution:

Area bounded by the two curves is image.

answered Oct 5, 2015 by Sammi Mentor
edited Oct 5, 2015 by Sammi
0 votes

2(a)

Step 1:

The functions are  image and image.

The area of the region bounded by the graphs of image and image and the vertical lines image and image is image.

Graph:

Graph the functions image and image.

image

Observe the graph:

The intersection points of the two curves are image and image.

image in the interval image.

Step 2:

Find the area of the region.

Substitute image, image and image in image.

image

Apply power rule of integration: image.

image

image

image

image

image

image.

Area bounded by the functions is image.

Solution:

Area bounded by the functions is image.

answered Oct 5, 2015 by Sammi Mentor
edited Oct 5, 2015 by Sammi
0 votes

2(b)

Step 1:

The curve equations are image and  image.

Graph the functions  image and image.

Shade the required region.

Observe the graph :

Region enclosed by the curves is between and .

Step 2:

Area between the curves and and between and is 

.

Here and .

Area of the region: .

Integral of symmetric function: If is even function, then .

Area of the region is sq-units.

Solution:

Area of the region is sq-units.

answered Oct 5, 2015 by cameron Mentor
edited Oct 5, 2015 by cameron
0 votes

(3)

Step 1:

The curve equations are , , the lines are and .

Graph the functions  , ,   and .

Shade the required region.

Observe the graph :

Region enclosed by the curves is between and .

Step 2:

Area between the curves and and between and is 

.

Observe the graph:

In the interval ,

Upper curve is and the lower curve is .

In the interval ,

Upper curve is and the lower curve is .

In the interval ,

Upper curve is and the lower curve is .

Area of the region: .

Area of the region is sq-units.

Solution:

Area of the region is sq-units.

answered Oct 5, 2015 by cameron Mentor
edited Oct 5, 2015 by cameron

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