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Find the area enclosed between the curves y = f(x) and y = g(x)

Don't forget, in order to find the area under a curve y=f(x) between two values x=a and x=b we integrate f(x) between a and b.Thus to find the area enclosed between two curves y=f(x) and y=g(x) we simply nee...
MG
Answered by Matthew G. • Maths tutor
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solve for x, in the form x = loga/logb for 2^(4x - 1) = 3^(5-2x) (taken from OCR June 2014 C2)

We can take logs of both sides straight away, and using the log rule that alogb = log(b^a)So(4x-1) log(2) = (5 - 2x) log(3)We can expand the bracketsso (4x)log(2) - log(2) = 5log(3) - 2xlog(3)We can group th...
EN
Answered by Ellie N. • Maths tutor
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The curve C has the equation y=((x^2+4)(x-3))/2*x where x is not equal to 0 . Find the tangent to the curve C at the point where x=-1 in the form y=mx+c

Firstly we need to expand out y into a series of terms to make it easier for us to compute the derivative . You multiply out the brackets to get y=(x 3 -3x 2 +4x-12)/2x then we divide each of the terms of th...
EM
Answered by Ellie M. • Maths tutor
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Let z=x+yi such that 16=5z - 3z*, What is z?

z* is the complex conjugate of z therefore z* = x - yi. So 16 + 32i = 5(x + yi)-3(x - yi), real: 16 = 5x - 3x => 16=2x => x=8, imaginary: 32 = 5y + 3y => 32 = 8y => y=4, therefore z = 8 + 4i.
BC
Answered by Ben C. • Maths tutor
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f(x)=12x^2e^2x - 14, find the x-coordinates of the turning points.

f(x)=(12x^2)(e^2x) - 14, so using the chain rule f'(x)=(24x)(e^2x) + (12x^2)(2e^2x).To find the turning points set f'(x)=0, so (24x)(e^2x) + (24x^2)(e^2x) = 0. Thus (24xe^2x)(1+x)=0. Thus x=0 or x=-1.
CH
Answered by Charlotte H. • Maths tutor
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