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We can see that the function is a sum of three terms so we can deal with each term separately and add them up. The term 8x³ and 5 are relatively straightforward and follow the standard rules fo...

dx=du/6 => (u-5)/6=x So the integral is now (2((u-5)/6)-3)(u^1/2) du/6 Which through simplifying becomes (1/36)(2u-28)(u^1/2)du = (1/36)(2u^3/2 -28u^1/2)du After integrating becomes (1/36)(4(u^5/2)/5 -...

The best method to revise is through solving past questions and timed exam papers. As long as you have your notes and understand all the basic principles, then you just need to practice in order to avoid ...

Let $ denote the integral symbol, as I am limited here by my keyboard.

Recall the formula for integration by parts:

$ u.(dv/dx) dx = uv - $ u(dv/dx) dx

So to find $^pi_0...

We shall differentiate each term in the equation with respect to x.

dy/dx (x²) = 2x

dy/dx (2y²) = 4y dy/dx

dy/dx (3x) = 3

So we now have the equation 2x +...