Question #276510

Calculate the area under the curve 𝑦 = 𝑥3 + 4𝑥 + 1 from x=-3 to x=3


1
Expert's answer
2021-12-13T16:55:09-0500

x-intercept: x=0.246x=-0.246

area = area under negative part of f(x) + area under positive part of f(x):


S=30.246(𝑥3+4𝑥+1)dx+0.2463(𝑥3+4𝑥+1)dx=S=-\int ^{-0.246}_{-3} (𝑥^3 + 4𝑥 + 1)dx+\int ^{3}_{-0.246} (𝑥^3 + 4𝑥 + 1)dx=


=(x4/4+2x2+x)30.246+(x4/4+2x2+x)0.2463==-(x^4/4+2x^2+x)|^{-0.246}_{-3}+(x^4/4+2x^2+x)|^{3}_{-0.246}=


=20.25+1830.12+0.246+20.25+18+30.12+0.246=76.75=20.25+18-3-0.12+0.246+20.25+18+3-0.12+0.246=76.75


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