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f(x)+f(x+1/2) =1f(x)=1-f(x+1/2)0∫2f(x)dx=0∫21-f(x+1/2)dx0∫2f(x)dx=2-0∫2f(x+1/2)dxtake (x+1/2)=vdx=dv0∫2f(v)dv=2-0∫2f(v)dv2(0∫2f(v)dv)=20∫2f(v)dv=10∫2f(x)dx=1
Given A and B A = | 1 0 1 | B = | 1 1 0 | | 1 1 0 | | 0 1 1 | | 0
find the area bounded by the curve y=5x^2-4x+3 from the limit x=0 to x=5
Given that 2t 2 y′′ + ty′ - 3 y = 0 Show that this given solution are form a fundamental set of solutions for the differential equation? Solution The two solutions f
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The Definite Integral Area under a Curve If there exists an irregularly shaped curve, y = f(x) then there is no formula to find out
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Horizontal asymptotes : Such as we can have vertical asymptotes defined in terms of limits we can also have horizontal asymptotes explained in terms of limits. Definition
Consider the following linear equations. x1-3x2+x3+x4-x5=8 -2x1+6x2+x3-2x4-4x5=-1 3x1-9x2+8x3+4x4-13x5=49
Example of Trig Substitutions Evaluate the subsequent integral. ∫ √((25x 2 - 4) / x) (dx) Solution In this type of case the substitution u = 25x 2 - 4 will not wo
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