ianb
- 17
- 0
simple integral -- but a question of methodology
\int \frac{(x-1)}{2} dx
First, try substitution. Let u = x - 1, and du = dx\int \frac{u}{2} du
\frac{1}{2} \int u \ du
\frac{1}{2} \times \frac{u^{2}}{2}
\frac{1}{2} \times \frac{(x-1)^2}{2}
\frac{(x-1)^{2}}{4} + C
Or we could do this:
\int \frac{(x-1)}{2} dx
\int \frac{1}{2}(x-1)dx
\frac{1}{2} \int(x-1)dx
\frac{1}{2} (\frac{x^2}{2} - x) + C
\frac{1}{4} (x^2 - 2x) + C
Why should one of them be right?
\int \frac{(x-1)}{2} dx
First, try substitution. Let u = x - 1, and du = dx\int \frac{u}{2} du
\frac{1}{2} \int u \ du
\frac{1}{2} \times \frac{u^{2}}{2}
\frac{1}{2} \times \frac{(x-1)^2}{2}
\frac{(x-1)^{2}}{4} + C
Or we could do this:
\int \frac{(x-1)}{2} dx
\int \frac{1}{2}(x-1)dx
\frac{1}{2} \int(x-1)dx
\frac{1}{2} (\frac{x^2}{2} - x) + C
\frac{1}{4} (x^2 - 2x) + C
Why should one of them be right?
Last edited: