Question

Seja f: R⇒R definida por f(x)=mx²+n ,tal que $\int\limits^1_0 {f(x)} \, dx$=7/3 e $\int\limits^2_1 {f(x)} \, dx$= 31/3 . Determine os valores de m e n.

Answer 1

$\int _{0}^{1} f(x)dx = \frac{7}{3} \\ \int _ {1}^{2} f(x)dx = \frac{31}{3}$

portanto

mx³/3+nx/{0}^{1}=7/3

m/3+n=7/3

mx³/3+nx/{1}^{2}=31/3

8m/3+2n-m/3-n=31/3

7m/3+n=31/3

n=7-m/3

7m+7-m/3=31/3

6m+7=31

6m=24

[m=4]

[n=1] //.