WAEC - Further Mathematics (2023 - No. 26)
Evaluate: \(\int(2x + 1)^3 dx\)
\(8(2x + 1)^2 + k\)
\(6(2x + 1)^2 + k\)
\(\frac{1}{8} (2x + 1)^4 + k\)
\(\frac{1}{6} (2x + 1)^4 + k\)
Објашњење
\(\int(2x + 1)^3 dx\)
Using substitution method, Let \(u = 2x + 1\)
\(\frac{du}{dx}=2==>du=2dx==>dx=\frac{du}{2}\)
=\(\int\frac{u^3}{2} du = \frac{1}{2}\int u^3 du\)
=\(\frac{1}{2}(\frac{u^4}{4})=\frac{u^4}{8}\)
\(\therefore\frac{1}{8} (2x + 1)^4 + k\)