Question

вычеслите интеграл ¹_3f(7x^2+2x-3)dx​

Answer 1

$\displaystyle\int\limits^1_{-3} {(7x^2+2x-3} )\, dx=(7\frac{x^3}{3}+2\frac{x^2}{2}-3x) \bigg|^1_{-3}=\\\\=(\frac{7}{3}*1^3+1^2-3*1)-(\frac{7}{3}*(-3)^3+(-3)^2-3*(-3))=\\\\=\frac{7}{3}+1-3+63-9-9= \frac{7}{3}+ 43=45\frac{1}{3}$