Answer:
The answer is " [tex]\bold{\frac{7}{2}}[/tex]"
Step-by-step explanation:
Given value:
[tex]\to f(x)=4x+3x^2-x^3\\\\\to g(x)=0[/tex]
Find:
area=?
calculation:
[tex]\ Area = \int_{-1}^{0} -(4x+3x^2-x^3) dx + \int_{0}^{1} (4x+3x^2-x^3) dx \\\\[/tex]
[tex]= -(2x^2+x^3- \frac{x^4}{4})_{-1}^{0} + (2x^2+x^3- \frac{x^4}{4})_{0}^{1}\\\\\\= -( 2x^2+x^3- \frac{x^4}{4})_{-1}^{0} + (2x^2+x^3- \frac{x^4}{4})_{0}^{1}\\\\\\= ( 2- 1 -\frac{1}{4}) + (2+1- \frac{1}{4})\\\\\\= ( \frac{8-4-1}{4}) + (\frac{8+4-1}{4})\\\\= ( \frac{3}{4}) + (\frac{11}{4})\\\\= \frac{3}{4} + \frac{11}{4}\\\\= \frac{11+3}{4}\\\\= \frac{14}{4}\\\\= \frac{7}{2}\\\\[/tex]
