We'll use the next basic integration formula:
∫xndx=n+1xn+1+C.
So we have:
1)
∫2xxdx=∫2x1x21dx=∫2x23dx=
=223+1x23+1+C=225x25+C=54x25+C=
=54x2+21+C=54x2x21+C=54x2x+C.
2) ∫x32dx=∫x232dx=∫2x−23dx=
=2−23+1x−23+1+C=2−21x−21+C=−x214+C=−x4+C.
Answer: 1) ∫2xxdx=54x2x+C,
2) ∫x32dx=−x4+C.
Comments
Leave a comment