# Thread: Help needed for intergration.

1. ## 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

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.

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

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

3. 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

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

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

3. int{(1/2)x^2 - 3x^1/2 - 5x^-1/3}dx

= (1/6)x^3 - 2x^(3/2) - (15/2)X^(2/3) + C ............by the power rule

6. don't be overwhelmed, the questions are not that involved, i just wanted to explain the process while i'm doing it. it would be good to practice a lot more problems like these. Now that you know how to do them, post questions and your solutions so someone here can give you feedback on what (if anything) you are doing wrong