Answer to Question #275031 in Calculus for Asif

Question

5. a) Calculate the area under the curve y = x ^ 3 + 4x + 1 from x = - 3 to x = 3 .

b) Evaluate: integrate x * e ^ x dx from 0 to 4

Answer 1

(a) $y = x ^ 3 + 4x + 1$

Graph will intersect the curve at $x=-0.25$

Area under the curve (green region) = $\int_{-0.25}^{3}(x^3+4x+1)dx-\int_{-3}^{-0.25}(x^3+4x+1)dx$

$(x^4/4+4x^2/2+x)_{-0.25}^{3} \ - (x^4/4+4x^2/2+x)_{-3}^{-0.25} \\ =27.89+21.13+3.25-(14.29+15.13+2.75)\\ =20.1$

(b) $\int_{0}^{4} xe^xdx= [xe^x-\int (e^x.1)dx]_{0}^{4}\\$

$=[xe^x-\int (e^x.1)dx]_{0}^{4}=[xe^x-e^x]_{0}^{4}=4.e^4-e^4-0+1=54.59$