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Figūra yra ribojama parabolės y = x2 + 1 ir tiesės y = ax + 1; čia a > 0. Su kuria a reikšme šios figūros plotas lygus 36 ?

Sprendimas:
Randame grafikų susikirtimo taškus:
x^2+1 =
a* x+1
x^2+1 = a* x+1$$x^{2}+1$$ = $$a\cdot x+1$$

x^2- a* x+1 = 1$$x^{2}-a\cdot x+1$$ = $$1$$

x^2- a* x-1+1 = 0$$x^{2}-a\cdot x-1+1$$ = $$0$$

x^2- a* x+0 = 0$$x^{2}-a\cdot x+0$$ = $$0$$

x^2- a* x = 0$$x^{2}-a\cdot x$$ = $$0$$

x = 0$$x$$ = $$0$$

x-a = 0$$x-a$$ = $$0$$

x = 0+a$$x$$ = $$0+a$$

x = a$$x$$ = $$a$$

Grafikai kertasi taškuose x = 0 ir x = a.
Norint rasti figūros plotą, reikia rasti tiesės ir parabolės skirtumo integralą nuo 0 iki a.
∫(0;a; a* x+1-( x^2+1)) =
36
∫(0;a; a* x+1-( x^2+1)) = 36$$\int_{0}^{a} (a\cdot x+1-(x^{2}+1))$$ = $$36$$

∫(0;a; a* x+1- x^2-1) = 36$$\int_{0}^{a} (a\cdot x+1-x^{2}-1)$$ = $$36$$

∫(0;a; a* x- x^2+1-1) = 36$$\int_{0}^{a} (a\cdot x-x^{2}+1-1)$$ = $$36$$

∫(0;a; a* x- x^2+0) = 36$$\int_{0}^{a} (a\cdot x-x^{2}+0)$$ = $$36$$

∫(0;a; a* x- x^2) = 36$$\int_{0}^{a} (a\cdot x-x^{2})$$ = $$36$$

( * a^2- * a^3) = 36$$(\frac{a}{2}\cdot a^{2}-\frac{1}{3}\cdot a^{3})$$ = $$36$$ 
( - * a^3) = 36$$(\frac{a^{3}}{2}-\frac{1}{3}\cdot a^{3})$$ = $$36$$ 
( - ) = 36$$(\frac{a^{3}}{2}-\frac{a^{3}}{3})$$ = $$36$$ 
- = 36$$\frac{a^{3}}{2}-\frac{a^{3}}{3}$$ = $$36$$ 
= 36$$\frac{a^{3}}{6}$$ = $$36$$ 
= 6* 6$$\frac{a^{3}}{6}$$ = $$6\cdot 6$$ 
a^3 = 6* 6* 6$$a^{3}$$ = $$6\cdot 6\cdot 6$$

a = saknis(3,
6* 6* 6)$$a$$ = $$\sqrt[3]{6\cdot 6\cdot 6}$$ 
a = 6$$a$$ = $$6$$

Atsakymas: a = 6
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