ƒ(x)=(sec(πx/4)tan(πx/4)) g(x)=(√2 - 4)x+4, x=0?
Find the area of the region (Calculus 2):
ƒ(x)=(sec(πx/4)tan(πx/4)) g(x)=(√2 - 4)x+4, x=0
Are you Einstein?
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Find the area of the region (Calculus 2):
ƒ(x)=(sec(πx/4)tan(πx/4)) g(x)=(√2 - 4)x+4, x=0
Are you Einstein?
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If I try to find the area of the region between the two curves and the x-axis, then A = ∫ {0, 1} sec (πx/4) tan (πx/4) dx + ∫ {1, 1.547) ((√2 - 4)x + 4) dx = sec (πx/4)] {0, 1] + ((√2 - 4)x^2/2 + 4x)] {1, 1.547} ≈ 0.914