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Use integration by parts to evaluate: ∫xsin(x) dx.

Since our function is a product of two "mini-functions" of x, we are able to use integration by parts.The trick for this is to correctly set 'u' and 'dv'. 'u' should be labelled as the function whi...
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Answered by Bailey A. • Maths tutor
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Find the area bounded be the curve with the equation y = x^2, the line x = 1, the line x = -1, and the x-axis.

The answer is 2/3. This can either be obtained by performing a standard integration of y=x^2, using the power rule, between x = 1 and x = -1. Alternatively, integrate y = x^2 between x = 0 and x = 1, then do...
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Answered by Isaac A. • Maths tutor
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How do you sketch the curve y=(x^2 - 4)(x+3), marking on turning points and values at which it crosses the x axis

First, factorise the equation into y = (x+3)(x-2)(x+2), nothing that x 2 -4 is the difference of two squares so is easy to factorise. From the factorised equation, the points at which the curve crosses the x...
Answered by • Maths tutor
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The sum of the first K natural numbers is 300. Find the value of K.

S= 0.5n(n+1) ( formula for the sum of the first n natural numbers) 0.5K(K+1)= 300 ( sub in K and make the formula equal to the total sum given) 0.5K(K+1)-300= 0 ( take away 300 from both sides) 0.5K 2 + 0.5K...
MB
Answered by Meghan B. • Maths tutor
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The function f(x)=x^2 -2x -24x^(1/2) has one stationary point. Find the value of x when f(x) is stationary, and hence determine the nature of this stationary point.

dy/dx = 2x - 2 -12x -1/2 dy/dx = 02x - 2 - 12x -1/2 = 0x = 4 (stationary point)d 2 y/dx 2 = 2 + 6x -3/2 when x=4, d 2 y/dx 2 = 11/411/4 > 0 therefore the point is a minimum
Answered by • Maths tutor
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