Solution assignment 04 Integration by substitution

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Assignment 4

Solve:

\displaystyle\int_{2}^{3}{\displaystyle\frac{1}{3x+1}}dx

Solution

We use the transform:

u=3x+1

and get:

\displaystyle\frac{du}{dx}=3

or:

dx=\displaystyle\frac{1}{3}du

We are not ready yet. There are boundaries from x=2 to x=3. These boundaries hold for the variable x and thus they have to be transformed as well in u. Thus the transformed boundaries are from u=7 to u=10. As a result we get the following integral:

\displaystyle\int_{7}^{10}{\displaystyle\frac{1}{u}\cdot\displaystyle\frac{1}{3}}du=\displaystyle\frac{1}{3}[\ln(|u|)]_{7}^{10}=\displaystyle\frac{1}{3}[\ln(10)-\ln(7)]=\displaystyle\frac{1}{3}\ln(\displaystyle\frac{10}{7})

We can verify whether the primitive function is correct by differentiation. The chain rule has to be applied.

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