Calculate the area under the curve 𝑦 = 𝑥3 + 4𝑥 + 1 from x=-3 to x=3
x-intercept: x=−0.246x=-0.246x=−0.246
area = area under negative part of f(x) + area under positive part of f(x):
S=−∫−3−0.246(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=S=−∫−3−0.246(x3+4x+1)dx+∫−0.2463(x3+4x+1)dx=
=−(x4/4+2x2+x)∣−3−0.246+(x4/4+2x2+x)∣−0.2463==-(x^4/4+2x^2+x)|^{-0.246}_{-3}+(x^4/4+2x^2+x)|^{3}_{-0.246}==−(x4/4+2x2+x)∣−3−0.246+(x4/4+2x2+x)∣−0.2463=
=20.25+18−3−0.12+0.246+20.25+18+3−0.12+0.246=76.75=20.25+18-3-0.12+0.246+20.25+18+3-0.12+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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