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a fost răspuns

Aflati aria subgraficului functiei:
a) f:[0, 1]→ R, f(x) = 4x( la 2) + 3x + 1

b) f:[0, pi/4]→ R, f(x) = sin2x

Răspuns :

a)A=[tex]4 \frac{x^3}{3}+3 \frac{ x^{2} }{2} +x=0-( \frac{4}{3}+ \frac{3}{2}+1)= \frac{8+9+6}{6}= \frac{23}{6} \\ b)A=- \frac{1}{2}cos2x= \frac{1}{2}*0+ \frac{1}{2}*1= \frac{1}{2} [/tex]

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