Объяснение:
∫(2х⁵+х²-7х+25)d(x)=х⁶/6+х³/3-7х²/2+25х+С
∫₋₂⁰(2х⁵+х²-7х+25)d(x)=(х⁶/6+х³/3-7х²/2+25х)⁰₋₂
=(0)-(64/6-8/3-28/2-50)=-32/3+8/3+14+50=56
-20∫₋₂⁰(4х³+2х)d(x)=-20(4х⁴/4+2х²/2)⁰₋₂=-20(х⁴+х²)⁰₋₂=-20((0+0)-(16+4))=400
Объяснение:
∫(2х⁵+х²-7х+25)d(x)=х⁶/6+х³/3-7х²/2+25х+С
∫₋₂⁰(2х⁵+х²-7х+25)d(x)=(х⁶/6+х³/3-7х²/2+25х)⁰₋₂
=(0)-(64/6-8/3-28/2-50)=-32/3+8/3+14+50=56
-20∫₋₂⁰(4х³+2х)d(x)=-20(4х⁴/4+2х²/2)⁰₋₂=-20(х⁴+х²)⁰₋₂=-20((0+0)-(16+4))=400