1. ## Random Integration Problems

2. For the q. attached, why can't a = -1 too?

3. Calculate the area bounded by the graphs of f(x) = x^2, g(x) = 1/x^2, x>0 and the line x =3. Could someone please verify the answer for me? The txtbk says it's 8 units squared but for some reason I only get 1 ><

Thanx a lot.

2. Originally Posted by xwrathbringerx
1. Show, by integration, that the area of a unit square is divided into four equal parts by the curves y = x^3, y = sqrt(x) and y = x. Could someone show me the correct way to do this proof?

2. For the q. attached, why can't a = -1 too?

Thanx a lot.
2. is easy.

$\int_{-1}^a{x\,dx} = 0$

$\left[\frac{1}{2}x^2\right]_{-1}^a = 0$

$\frac{1}{2}[a - (-1)] = 0$

$\frac{1}{2}(a + 1) = 0$

$a + 1 = 0$

$a = -1$.

3. Hmmm I get both 1 AND - 1. The txtbk answer says it's only 1 though...

4. Corrections in red:

Originally Posted by Prove It
2. is easy.

$\int_{-1}^a{x\,dx} = 0$

$\left[\frac{1}{2}x^2\right]_{-1}^a = 0$

$\frac{1}{2} [a{\color{red}^2} - (-1){\color{red}^2}] = 0$

$\frac{1}{2}(a{\color{red}^2 - } 1) = 0$

$a{\color{red}^2 -} 1 = 0$

$a = {\color{red}\pm} 1$.
@OP: a = -1 is a perfectly good (if trivial) solution. Perhaps the book only wanted NON-trivial solutions.

5. Originally Posted by mr fantastic
Corrections in red:

@OP: a = -1 is a perfectly good (if trivial) solution. Perhaps the book only wanted NON-trivial solutions.
OMG my solutions have been shocking today... I must need sleep...

Thanks Mr F.

6. Originally Posted by xwrathbringerx
[snip]
3. Calculate the area bounded by the graphs of f(x) = x^2, g(x) = 1/x^2, x>0 and the line x =3. Could someone please verify the answer for me? The txtbk says it's 8 units squared but for some reason I only get 1 ><

Thanx a lot.
Area = $\int_1^3 x^2 - \frac{1}{x^2} \, dx = 8$.

7. Ummm for q. 3, is this the diagram for the area or is it the shaded area onli ABOVE the x-axis?

8. Originally Posted by xwrathbringerx
Ummm for q. 3, is this the diagram for the area or is it the shaded area onli ABOVE the x-axis?
No. Your graphs are wrong. Try plotting points. Note that:

f(1) = g(1) = 1.

f(3) = 9 and g(3) = 1/9.