Help with area under curve?

[ventaak]

Well I've gone back to college after a looooooong hiatus and they stuck me in Calculus II despite telling them I haven't touched a math textbook since 2008. It's only been two classes and I'm already banging my head against the wall.

Anyway, can someone tell me what was used to get the second line, (11/4 - X^2 - X)?

Example: The area under the graph of \(\displaystyle \, f(x)\, =\, 3\, -\, \left(x\, +\, \dfrac{1}{2}\right)^2\, \) for \(\displaystyle \, x\, \in\, [-1,\, 1]\, \) is determined as follows:

. . .\(\displaystyle \displaystyle A\, =\, \int_{-1}^1\, 3\, -\, \left(x\, +\, \dfrac{1}{2}\right)^2\, dx\)

. . . . .\(\displaystyle \displaystyle =\, \int_{-1}^1\, \dfrac{11}{4}\, -\, x^2\, -\, x\, dx\)

. . . . .\(\displaystyle =\, \dfrac{11}{4}\, x\, -\, \dfrac{1}{3}\, x^3\, -\, \dfrac{1}{2}\, x^2\, \Bigg\rvert_{-1}^1\)

. . . . .\(\displaystyle =\, \left(\dfrac{11}{4}\, -\, \dfrac{1}{3}\, -\, \dfrac{1}{2}\right)\, -\, \left(-\dfrac{11}{4}\, +\, \dfrac{1}{3}\, -\, \dfrac{1}{2}\right)\)

. . . . .\(\displaystyle =\,\dfrac{11}{2}\, -\, \dfrac{2}{3}\, =\, \dfrac{33}{6}\, -\, \dfrac{4}{6}\, =\, \dfrac{29}{6}\)

. . . . .\(\displaystyle =\,4.8\overline{3}\)

Was it simply Algebra (in which case I've must of forgotten way more thought) or is it some sort of anti-differentiation rule?

 

[Deleted member 4993]

Well I've gone back to college after a looooooong hiatus and they stuck me in Calculus II despite telling them I haven't touched a math textbook since 2008. It's only been two classes and I'm already banging my head against the wall.

Anyway, can someone tell me what was used to get the second line, (11/4 - X^2 - X)?

Example: The area under the graph of \(\displaystyle \, f(x)\, =\, 3\, -\, \left(x\, +\, \dfrac{1}{2}\right)^2\, \) for \(\displaystyle \, x\, \in\, [-1,\, 1]\, \) is determined as follows:

. . .\(\displaystyle \displaystyle A\, =\, \int_{-1}^1\, 3\, -\, \left(x\, +\, \dfrac{1}{2}\right)^2\, dx\)

. . . . .\(\displaystyle \displaystyle =\, \int_{-1}^1\, \dfrac{11}{4}\, -\, x^2\, -\, x\, dx\)

. . . . .\(\displaystyle =\, \dfrac{11}{4}\, x\, -\, \dfrac{1}{3}\, x^3\, -\, \dfrac{1}{2}\, x^2\, \Bigg\rvert_{-1}^1\)

. . . . .\(\displaystyle =\, \left(\dfrac{11}{4}\, -\, \dfrac{1}{3}\, -\, \dfrac{1}{2}\right)\, -\, \left(-\dfrac{11}{4}\, +\, \dfrac{1}{3}\, -\, \dfrac{1}{2}\right)\)

. . . . .\(\displaystyle =\,\dfrac{11}{2}\, -\, \dfrac{2}{3}\, =\, \dfrac{33}{6}\, -\, \dfrac{4}{6}\, =\, \dfrac{29}{6}\)

. . . . .\(\displaystyle =\,4.8\overline{3}\)

Was it simply Algebra (in which case I've must of forgotten way more thought) or is it some sort of anti-differentiation rule?

(a+b)² = a² + 2ab + b²

(x + 1/2)² = ??

 

[stapel]

Can someone tell me what was used to get the second line, (11/4 - X^2 - X)?

Example: The area under the graph of \(\displaystyle \, f(x)\, =\, 3\, -\, \left(x\, +\, \dfrac{1}{2}\right)^2\, \) for \(\displaystyle \, x\, \in\, [-1,\, 1]\, \) is determined as follows:

. . .\(\displaystyle \displaystyle A\, =\, \int_{-1}^1\, 3\, -\, \left(x\, +\, \dfrac{1}{2}\right)^2\, dx\)

. . . . .\(\displaystyle \displaystyle =\, \int_{-1}^1\, \dfrac{11}{4}\, -\, x^2\, -\, x\, dx\)

They skipped some steps. You can figure this out by doing the algebra and arithmetic yourself (or, which is sometimes the case, you can do the algebra and figure out that you're missing something else, because the algebra doesn't get you all the way there):

. . . . .\(\displaystyle \left(x\, +\, \dfrac{1}{2}\right)^2\, =\, x^2\, +\, 2\left(\dfrac{1}{2}\right)(x)\, +\, \left(\dfrac{1}{2}\right)^2\, =\, x^2\, +\, x\, +\, \dfrac{1}{4}\)

. . . . .\(\displaystyle 3\, -\, \left(x\, +\, \dfrac{1}{2}\right)^2\, =\, 3\, -\, \left(x^2\, +\, x\, +\, \dfrac{1}{4}\right)\, =\, 3\, -\, x^2\, -\, x\, -\, \dfrac{1}{4}\, =\, 3\, -\, \dfrac{1}{4}\, -\, x^2\, -\, x\)

What do you get as your next step?

 

[ventaak]

I get it now. Thanks for the reply. I've got a lot of reviewing to do. This is going to be a long semester. =/