Answer to Question #276510 in Calculus for RIYAD

Question

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

Answer 1

x-intercept: $x=-0.246$

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

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

$=-(x^4/4+2x^2+x)|^{-0.246}_{-3}+(x^4/4+2x^2+x)|^{3}_{-0.246}=$

$=20.25+18-3-0.12+0.246+20.25+18+3-0.12+0.246=76.75$