Question

evaluate the definite integral

[-1,1] (x^3+bx^2)dx

Answer 1

int (x^3+bx^2)dx

(1/4)x^4 + (b/3)x^3 + C


Now evaluate limits of integration


((1/4)(1)^4 + (b/3)(1)^3) - ((1/4)(-1)^4 + (b/3)(-1)^3)


(1/4 + b/3) - (1/4 - b/3)


1/4 + b/3) - 1/4 + b/3


2b/3


So the entire expression evaluates to 2b/3

Answer 2

\[\int_{-1}^{1} x^3+bx^2\ dx=\int_{-1}^{1} x^3\ dx+b\int_{-1}^1x^2\ dx\]
Note that x^3 is odd while x^2 is even.