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 $\displaystyle \displaystyle \int_0^2{\mid{6x - 7}\mid dx$.

What I have done is:

-evaluated the function

a) $\displaystyle 6x - 7 < 0 for 0\leq x < \frac{7}{6}$ and $\displaystyle 6x - 7 > 0 for \frac{7}{6} < x \leq 2$

we have $\displaystyle {\mid{6x - 7}\mid = -(6x-7) = 7-6x, for 0\leq x < \frac{7}{6}$ and $\displaystyle {\mid{6x - 7}\mid = (6x-7) for \frac{7}{6} < x \leq 2 $

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

$\displaystyle \displaystyle \int_0^2{\mid{6x - 7}\mid dx = \displaystyle \int_{0}^{\frac{7}{6}}(7 - 6x)dx + \displaystyle \int_{\frac{7}{6}}^{2}(6x - 7)dx$

$\displaystyle \displaystyle (7x - 3x^2)\right|_{0}^{\frac{7}{6}} + (3x^2 - 7x)\right|_{\frac{7}{6}}^{2}$

$\displaystyle [ (\frac{49}{6} - \frac{147}{6}) - (0)] + [(12-14) - (\frac{147}{6} - \frac{49}{6})] = (-\frac{98}{6}) + (- \frac{110}{6}) = - \frac{208}{6}$ ,

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

$\displaystyle \displaystyle (7x - 3x^2)\right|_{0}^{\frac{7}{6}} + (3x^2 - 7x)\right|_{\frac{7}{6}}^{2}$

$\displaystyle [ (\frac{49}{6} - \frac{147}{36}) - (0)] + [(12-14) - (\frac{147}{36} - \frac{49}{6})] = 2$

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 $\displaystyle 6x - 7 |_{x = 2}$.

Re: Evaluate the integral

Quote:

Originally Posted by

**FelixFelicis28** The area of triangle 2 is incorrect. Your base is correct but your height should $\displaystyle 6x - 7 |_{x = 2}$.

(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?

$\displaystyle \frac{74}{12} = \frac{37}{6}$ is correct. :)