Friday

January 30, 2015

January 30, 2015

Posted by **Annie** on Thursday, January 5, 2012 at 9:44pm.

- calculus -
**Steve**, Friday, January 6, 2012 at 11:02amy = √x

y' = 1/2√x

So, at (4,2) the tangent line has slope 1/4.

So, the equation for the line is

(y-2)/(x-4) = 1/4

y = x/4 + 1

which insersects tha x-axis at x = -4.

Now, √x is not defined for x<0, so we need to break the region into two parts:

Int(x/4 + 1)dx [-4,0] + Int(x/4 + 1 - √x)dx [0,4]

= [x^2/8 + x)[-4,0] + (x^2/8 + x - 2/3x^3/2)[0,4]

8 - 16/3 = 8/3

**Answer this Question**

**Related Questions**

Algebra - I need some help figuring these. Compute each product and simplify ...

Algebra - I need some help figuring these. Compute each product and simplify the...

calculus - write the expression in the form bi, where b is the real number ...

math - Classify each statement as sometimes, always, or never true. A. 2^(2/3)*4...

algebra - ^5 squareroot of x^20 *my answer x^4 ^3 squareroot of 729/343 my ...

math 11 review - A couple of extra practice questions I can't remember how to do...

Calculus 1 - Let f be the function given by f(x)=3sqrt(x-2). A) On the axes ...

Algebra2 - Simplify: 1.SquareRoot(-144) 2.SquareRoot(-64x^4) 3.SquareRoot(-13)*...

calculus - find the area enclosed by the given curves. y=squareroot of x y=x/2 ...

Algebra - I am not too sure about this problem. Can you please help me? Graph ...