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Since we differentiate a function to find the gradient of a curve at any point, we need to reverse that to find the equation of the curve. We do this by integrating with respect to x:If you have a constan...
Formula for a straight line y-y1=m(x-x1), where m is the gradient substituting in the values given to find the gradient we get 4-1=m(3+2), therefore m= 3/5the midpoint of the two poi...
When a function is increasing, it’s derivative is positive. So first let us differentiate f(x). To differentiate xn we multiply by n and then reduce the power by 1. So f’(x)=8-2*2x=8-4x. We wan...
R=sqrt(42)=6.48...a=arcos(4/sqrt(42))=0.905...Rsin(x+a)=6.48sin(x+0.905)
Take ln(x) = 1 * ln(x) and integrate by parts. Let u = ln(x) and dv/dx = 1 such that du/dx = 1/x and v = x. Using the integrating by parts formula, you get x ln(x) - integral(1) = x ln(x) - x + c.
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