Calculus and Differentiation / Intergration Questions?

Question

1) Use calculus to determine an approximate value for 2 / cube root of 7.8, leaving your answer correct to 4 decimal places.

2) Differentiate sqrt (x^2 + 9) with respect to x and hence find the exact value of the intergration of (3x) / sqrt (2x^2 + 18) with limits 0 to 4.

*Please show all essential workings. Thanks (:

Answer 1

1)we have to calculate 2/(7.8)^(1/3)
y = 2/x^(1/3)
dy = (-2/3) * x^(-4/3) dx
here x=8,y = 1 , dx = -0.2, we have to calculate dy
so, dy = -2/3 * 8^(-4/3) *(-0.2) = -2/3 * 1/16 *-0.2 = 0.0083
2/7.8^(1/3) = y+dy = 1+0.0083 = 1.0083

2)d(x^2+9) = d(x^2)+d(9) = 2x +0 (as d(x^n) = n*x^(n-1) and differentiation of constant is zero)

now for integration of (3x) / sqrt (2x^2 + 18)
let z = 2x^2 + 18
so, dz = 4x dx ; dz/4 = x dx
so we can write,
3x dx/sqrt (2x^2 + 18) = 3/4 dz / z^(1/2)= 3/4 * z^(-1/2) dz
now integration of z^n dz = z^(n+1) / (n+1)
so, integration of z^(-1/2) dz = z^(1/2)/(1/2) = 2* sqrt(z)
now change the limits for x = 0, z= 18 and for x=4,z= 50
value for z=50 , 2 * sqrt(50) = 10*sqrt(2)
value for z=18 , 2 * sqrt(18) = 6*sqrt(2)
so required value = upper value - lower value = 10*sqrt(2) - 6*sqrt(2) = 4*sqrt(2) = 4 * 1.414 = 5.656

Answer 2

1) dy/dx = Dy/Dx, where D = approximation.
So Dy = (dy/dx)(Dx) ...(1)

Let y = 2x^(-1/3)
so dy/dx = (-2/3)x^(-4/3) ...(2)

In order to find approximation Dy, so substitute x = 8 in (2) and Dx = -0.2.

Dy = 1/120

2/[cube root 7.8] = y + Dy = 1+(1/120) = 1.0083

2) y = sqrt(x^2+9)
y'= 2x/sqrt(x^2+9)

So,
Int(0,4) [3x/sqrt(2x^2+18)]dx
= [3/2sqrt(2)] {Int(0,4) [2x/sqrt(x^2+9)]dx}
= [3/2sqrt(2)] {sqrt(4^2+9)-sqrt(9)]
= 3/sqrt(2)

Answer 3

2) y´= x/sqrt(x^2+9)
Int(0,4) 3x/sqrt(2x^2+18)dx
put 2x^2+18 =z so 2xdx = dz
3/2Int z^-1/2 dz = 3 z^1/2 (0,50) = 3*(sqrt(50)-sqrt(18))
1) 7.8 = 8-0.2
= 8(1-0.025)
2/cube root(7.8) = (1-0.025) ^-1/3 = 1+1/3*0.025 -2/9(0.025)^2=
1.0084

Answer 4

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