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Find the area under the curve y = sin(2x) + cos(x) between 0 and pi/2

to find the area integrate the function between 0 and pi/2: indefinite integral of y = -1/2 cos(2x) + sin(x) + c ignore c and input boundary conditions: [-1/2 cos(pi) + sin(pi/2)] - [-1/2 cos(0) + sin(0)] = ...
HD
Answered by Harry D. • Maths tutor
8039 Views

Given that y = 4x^5 - 5/(x^2) , x=/=0 , find a)dy/dx b)indefinite integral of y

rewrite as y = 4x^5 - 5x^(-2) dy/dx = 20x^4 + 10x^(-3) integral of y = 4x^6 / 6 + 5x^(-1) / 1 + c simplify: 2/3 x^6 + 5/x
HD
Answered by Harry D. • Maths tutor
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Differentiate with respect to x: i) y=x^3ln(2x) ii) y=(x+sin(2x))^3

i) There first step is to acknowledge the need for both product rule (d(uv)/dx=v.du/dx+u.dv/dx) and chain rule (dz/dx=dz/dy*dy/dx). Here, u=x^3 and v=ln(2x). Therefore, du/dx=3x^2 which is a standard differe...
EM
Answered by Edward M. • Maths tutor
10092 Views

Differentiate f(x) = (x+3)/(2x-5) using the quotient rule.

For a quotient f(x) = u(x)/v(x), the derivative is f'(x) = (v u'(x) - u v'(x))/v(x) 2 . Applying this to the given function, we find u(x) = x+3 and v(x) = 2x-5. So, u'(x) = 1 and v'(x) = 2. We can then put t...
SR
Answered by Sara R. • Maths tutor
6551 Views

Integrate 2x/(x^2+3) using the substitution u=x^2+3

u=x 2 + 3 du/dx=2x dx=du/2x 2x/(x 2 +3) dx becomes (2x/u) * (du/2x) the 2x terms cancel out giving 1/u du this integrates to ln(u)+c becoming ln(x 2 +3)+c
TS
Answered by Tom S. • Maths tutor
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