Re: Evaluate the integral

Quote:

Originally Posted by

**dokrbb** So, here I am, beginning the calc II and beginning to make new mistakes as well ;), I have this problem, which states:

-Evaluate the integral by interpreting it in terms of areas. In other words, draw a picture of the region the integral represents, and find the area using high school geometry

of

.

What I have done is:

-evaluated the function

a)

and

we have

and

- then, by evaluating the integral, I have got:

,

and I must have made a mistake somewhere, because it say that it's wrong, or, maybe I should solve it by another method, but I have no idea how...

thank you, for all your input,

dokrbb

It should be

Re: Evaluate the integral

now, looking for another example I just understood that I have to do otherwise:

i have to draw a line corresponding to the function (6x-7) which intersects the x axis in the point 7/6, and I have 2 triangles I can evaluate as it asks with basics geometry;

since it asks me the absolute value, I have (A of triangle 1) + (A of triangle 2) = 1/2(7/6)(7) + 1/2(5/6)(2) = 49/12 + 10/12 = 59/12, but is considered also wrong, what don't I get here?

Re: Evaluate the integral

Quote:

Originally Posted by

**dokrbb** now, looking for another example I just understood that I have to do otherwise:

i have to draw a line corresponding to the function (6x-7) which intersects the x axis in the point 7/6, and I have 2 triangles I can evaluate as it asks with basics geometry;

since it asks me the absolute value, I have (A of triangle 1) + (A of triangle 2) = 1/2(7/6)(7) + 1/2(5/6)(2) = 49/12 + 10/12 = 59/12, but is considered also wrong, what don't I get here?

we were writing in the same time the posts, can you look at that now, thanks

Re: Evaluate the integral

Quote:

Originally Posted by

**dokrbb** now, looking for another example I just understood that I have to do otherwise:

i have to draw a line corresponding to the function (6x-7) which intersects the x axis in the point 7/6, and I have 2 triangles I can evaluate as it asks with basics geometry;

since it asks me the absolute value, I have (A of triangle 1) + (A of triangle 2) = 1/2(7/6)(7) + 1/2(5/6)(2) = 49/12 + 10/12 = 59/12, but is considered also wrong, what don't I get here?

The area of triangle 2 is incorrect. Your base is correct but your height should .

Re: Evaluate the integral

Quote:

Originally Posted by

**FelixFelicis28** The area of triangle 2 is incorrect. Your base is correct but your height should

.

(A of triangle 1) + (A of triangle 2) = 1/2(7/6)(7) + 1/2(5/6)**(5)** = 49/12 + **25/12 = 74/12,** so here is the mistake, right?

Re: Evaluate the integral

Quote:

Originally Posted by

**dokrbb** (A of triangle 1) + (A of triangle 2) = 1/2(7/6)(7) + 1/2(5/6)**(5)** = 49/12 + **25/12 = 74/12,** so here is the mistake, right?

is correct. :)