Question

Помогите пожалуйста, заранее благодарю.

Answer 1

$\displaystyle\\\int\limits^4_1 {3x^3+2x+5} \, dx =\bigg(\frac{3x^4}{4}+x^2+5x\bigg)\mid^4_1=\frac{3*4^4}{4}+4^2+5*4-(\frac{3*1^4}{4}+\\\\\\+1^2+5*1)=3*4^3+16+20-\bigg(\frac{3}{4}+6\bigg)=192+16+20-\frac{27}{4}=\\\\\\ =228-\frac{27}{4}=\frac{885}{4}\\\\\\ \int\limits^1_{-2} {\frac{4x^4+7x}{x} } \, dx = \int\limits^1_{-2}(4x^3+7)\ dx=(x^4+7x)\mid^1_{-2}=1^4+7*1-((-2)^4+\\\\\\+7*(-2))=1+7-(16-14)=8-(2)=8-2=6$