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Solve the simultaneous equations: y+4x+1=0 and y^2+5x^2+2x=0

y= -4x-1y 2 = (-4x-1) 2 = 16x 2 +8x +1y 2 +5x 2 +2x = 0lets substitute what we found y 2 equal to earlier, which gives us(16x 2 +8x +1) + 5x 2 +2x = 021x 2 + 10x +1 = 0after factorising this we get(7x+1)(3x+...
OU
Answered by Oleksandr U. Maths tutor
5147 Views

Solve int(ln(x)dx)

To solve this we must use integration by parts: int(udv) = uv - int(vdu) (1) Hence let u = ln(x), dv = dx => du=(1/x)dx, v=x, and now using (1) and substituting values we obtain int(ln(x)dx) = ln(x)x - in...
GB
Answered by George B. Maths tutor
3929 Views

What is integration?

Integration can be viewed in many ways. The most common way to interpret an integral is to take the area under the curve you would like to integrate. For example Draw y=x, limit between 0 and 1, shade in the...
MS
Answered by Mikhail S. Maths tutor
3435 Views

Solve: x^2 + y^2 = 25 y - 3x = 13

Equation 1) x 2 + y 2 = 25 Equation 2) y - 3x = 13 First you need to substitute a variable so there is only one unknown in the equation: 2) y = 13 + 3x Substituting this into equation 1) gives: x 2 + (13 + 3...
LW
Answered by Lucy W. Maths tutor
12051 Views

integrate cos^2(2x)sin^3(2x) dx

To integrate this we need to use the chain rule, substituting cos2x = u Integral becomes: u 2 sin 3 2x dxChain rule: dy/dx = du/dx dy/du du/dx = -2sin2x --> dx = -1/2sin2x du Substituting into the equatio...
LW
Answered by Lucy W. Maths tutor
7821 Views