# Help needed for intergration.

• Mar 6th 2007, 11:44 AM
Help needed for intergration.
Hi I need help for my homework. I dont know how to do X squared so X^2 means X squared and etc also I dont know how to do square root so root means square root. Also f means the elongated s:p

1. Find f(3X+ 5rootX) dX

2. find f(X-1)(2X+5)dx

3. find f 1/2X^2 - 3X^1/2 - 5X^-1/3 dx

4. find f (X+2)^2 / X^1/2 dx

5. given that dy/dx = 2X+X^-2 and that y=2 at x=1 find y in terms of X then the value of y at X=-1

6. The curve for which dy/dx = 2X - 1/rootX passes through the point (9,76). Find the equation of the curve in the form y=f(X) and then the value of y when x=4

7. The curve C has equation y=f(X) and the point P (2,7) lies on C. given that f ' (X) = 3X^2 - 4X + 5.
Find f(X) and then the point Q also lies on C, and the tangent at Q is parallel to the tangent to C at P. Find the co-ordinate of X for Q.

Sorry I asked for so much but I self teach and these are past paper questions that I don't understand, so sorry to cause trouble and if you dont mind can you also write out your workings as well to see how to come to that conclusion so I can handle questions of the smilar sort as future reference.

Thank You.
• Mar 6th 2007, 12:32 PM
Jhevon
Quote:

Hi I need help for my homework. I dont know how to do X squared so X^2 means X squared and etc also I dont know how to do square root so root means square root. Also f means the elongated s:p

1. Find f(3X+ 5rootX) dX

2. find f(X-1)(2X+5)dx

3. find f 1/2X^2 - 3X^1/2 - 5X^-1/3 dx

4. find f (X+2)^2 / X^1/2 dx

5. given that dy/dx = 2X+X^-2 and that y=2 at x=1 find y in terms of X then the value of y at X=-1

ok, so yes, for people like us who dont use LaTex, x^2 means x squared. personally i use sqrt() to mean squareroot, and int{}dx for integrate something with respect to x

1. Find int{3x+ 5sqrt(x)}dx

int{3x+ 5sqrt(x)}dx
= int{3x+ 5x^(1/2)}dx
= (3/2)x^2 + (10/3)x^(3/2) + C

remember, to integrate a function of the form x^n, it is int{x^n}dx = (x^(n+1))/n+1. as in, you add 1 to the power and divide by the new power, this is called the "power rule"

2. find int{(X-1)(2X+5)}dx

int{(X-1)(2X+5)}dx
= int{2x^2 + 3x - 5}dx .............expand the brackets
= (2/3)x^3 + (3/2)x^2 - 5x + C

3. find f 1/2X^2 - 3X^1/2 - 5X^-1/3 dx

is it (1/2)x^2 or 1/(2x^2)?

4. find f (X+2)^2 / X^1/2 dx

int{[(x+2)^2]/x^(1/2)}dx
= int{(x^2 + 4x + 4)/x^(1/2)}dx
= int{x^(3/2) + 4x^(1/2) + 4x^(-1/2)}dx .............do you understand this step? i divided each term of the top by x^(1/2)
= (2/5)x^(5/2) + (8/3)x^(3/2) + 8x^(1/2) + C

5. given that dy/dx = 2X+X^-2 and that y=2 at x=1 find y in terms of X then the value of y at X=-1

dy/dx means the derivative of y with respect to x. this means the function you see was differentiated, we have to integrate to get back to the original function.

dy/dx = 2x + x^-2
=> dy = (2x + x^-2)dx
=>int{1}dy = int{2x + x^-2}dx
=> y = x^2 - x^-1 + C

when y=2, x = 1
=> 2 = 1^2 - 1^-1 + C
=> 2 = C

so our original function is y = x^2 - 1/x + 2

now to find y at x = -1

y = (-1)^2 - 1/-1 + 2
y = 1 + 1 + 2 = 4
• Mar 6th 2007, 12:54 PM
wow thank you but I wonder if its possible if 6 and 7 is completed as well please.

sorry for question 3 its (1/2)x^2
• Mar 6th 2007, 01:04 PM
Jhevon
Quote:

6. The curve for which dy/dx = 2X - 1/rootX passes through the point (9,76). Find the equation of the curve in the form y=f(X) and then the value of y when x=4

7. The curve C has equation y=f(X) and the point P (2,7) lies on C. given that f ' (X) = 3X^2 - 4X + 5.
Find f(X) and then the point Q also lies on C, and the tangent at Q is parallel to the tangent to C at P. Find the co-ordinate of X for Q.

Sorry I asked for so much but I self teach and these are past paper questions that I don't understand, so sorry to cause trouble and if you dont mind can you also write out your workings as well to see how to come to that conclusion so I can handle questions of the smilar sort as future reference.

Thank You.

6. The curve for which dy/dx = 2X - 1/rootX passes through the point (9,76). Find the equation of the curve in the form y=f(X) and then the value of y when x=4

same principle as question 5

dy/dx = 2x - 1/sqrt(x)
=> y = int{2x - x^(-1/2)}dx ..........note, sqrt(x) = x^(1/2), so x^(-1/2) is 1/sqrt(x). we use negative exponents to mean the inverse, as in 1 over whatever the function is
=> y = x^2 - 2x^(1/2) + C
this curve passes through (9,76), that means when x = 9, y = 76.
=> 76 = (9)^2 - 2(9)^(1/2) + C
=> 76 = 81 - 6 + C
=> C = 1

so y = x^2 - 2x^(1/2) + 1

when x = 4

y = (4)^2 - 2(4)^(1/2) + 1 = 16 - 4 + 1 = 17 - 4 = 13

7. The curve C has equation y=f(X) and the point P (2,7) lies on C. given that f ' (X) = 3X^2 - 4X + 5.
Find f(X) and then the point Q also lies on C, and the tangent at Q is parallel to the tangent to C at P. Find the co-ordinate of X for Q.

f ' (x) also means the derivative of f(x), so it begins the same way as questions 5 and 6.

f'(x) = 3x^2 - 4x + 5
=> f(x) = int{3x^2 - 4x + 5}dx
=> f(x) = x^3 - 2x^2 + 5x + C

y passes through (2,7), so when x=2, y=7

=> 7 = (2)^3 - 2(2)^2 + 5(2) + C
=> 7 = 8 - 8 + 10 + C
=> C = 7 - 10 = -3

so y = x^3 - 2x^2 + 5x - 3

now for the second part, you must recall that the derivative gives the slope of the tangent line for any point of x. we can find the slope at P using the derivative and then find all x's that give that same slope

P(2,7) and f'(x) = 3X^2 - 4X + 5

=> slope of tangent line at P = 3(2)^2 - 4(2) + 5 = 12 - 8 + 5 = 9. ....plugged in the value of x at P

now since tangent line at Q is parallel to the tangent line at P, it has the same slope. so m for Q is 9 also.

we use the derivative to find all values of x where the slope is 9

we want therefore, f'(x) = 9
=> 3X^2 - 4X + 5 = 9
=> 3x^2 - 4x - 4 = 0
(3x + 2)(x - 2) = 0
=> x = -2/3 or x = 2
x = 2 at P, so that means x = -2/3 at Q
• Mar 6th 2007, 01:07 PM
Jhevon
Quote: