Výpočet obsahu mezi křivkami
Urči obsah mezi křivkami funkcí na daném intervalu:
\( \large f\left( x\right) = x^2+1\text{a}g\left( x\right) = 0,\text{na intervalu}\left \langle-1;2 \right \rangle \)
\( S=\large\int\limits_{-1}^{\Large2}\left(x^2+1\right)dx=\left\lbrack\frac{x^3}{3}+x\right\rbrack_{-1}^2=\frac{2^3}{3}+2-\left\lbrack\frac{\left(-1\right)^3}{3}-1\right\rbrack= \)
\( \large{\displaystyle=\Large\frac{8}{3}+2+\Large\frac{1}{3}+1=8j^2} \)
\( S=\large\int\limits_{-1}^{\Large2}\left(x^2+1\right)dx=\left\lbrack\frac{x^3}{3}+x\right\rbrack_{-1}^2=\frac{2^3}{3}+2-\left\lbrack\frac{\left(-1\right)^3}{3}-1\right\rbrack= \)
\( \large{\displaystyle=\Large\frac{8}{3}+2+\Large\frac{1}{3}+1=7j^2} \)
\( S=\large\int\limits_{-1}^{\Large2}\left(x^2+1\right)dx=\left\lbrack\frac{x^3}{3}+x\right\rbrack_{-1}^2=\frac{2^3}{3}+2-\left\lbrack\frac{\left(-1\right)^3}{3}-1\right\rbrack= \)
\( \large{\displaystyle=\Large\frac{8}{3}+2+\Large\frac{1}{3}+1=6j^2} \)
\( S=\large\int\limits_{-1}^{\Large2}\left(x^2+1\right)dx=\left\lbrack\frac{x^3}{3}+x\right\rbrack_{-1}^2=\frac{2^3}{3}+2-\left\lbrack\frac{\left(-1\right)^3}{3}-1\right\rbrack= \)
\( \large{\displaystyle=\Large\frac{8}{3}+2+\Large\frac{1}{3}+1=5j^2} \)