In Monday’s class, student found many antiderivatives using the Substitution Method. The basic process is illustrated below for the antiderivative
$latex \displaystyle \int{4{{\left( {{x}^{2}}-3 \right)}^{3}}\cdot 2x,dx}$.
Let’s look at the basic steps.
- Choose the expression for u. This is generally the inside part of a composition in the integrand. Use the derivative to find an other expression for du.
- Match the integrand with u and du. All variables in the original integrand must change to u.
- Change the integrand so that it is written in terms of u.
- Work out the antiderivative in terms of u.
- Put in the expression for u so that the antiderivative is written in terms of the original variable.
Now let’s look at the examples carried out in class.
Problem 1 –
$latex \displaystyle \int{{{\left( 3{{x}^{2}}+4 \right)}^{4}}\cdot 6x,dx}$
Problem 2 –
$latex \displaystyle \int{{{\left( 3{{x}^{2}}+4 \right)}^{3}}\cdot 4x,dx}$
Problem 3 –
$latex \displaystyle \int{{{\left( {{x}^{2}}-1 \right)}^{5}}\cdot x,dx}$
Problem 4 –
$latex \displaystyle \int{4{{\left( 2x+3 \right)}^{2}},dx}$
Problem 5 –
$latex \displaystyle \int \frac{2}{{{\left( 2m+1 \right)}^{3}}} dm$
Problem 6 –
$latex \displaystyle \int{\frac{3}{\sqrt{3u-5}},du}$
Problem 7 –
$latex \displaystyle \int{-4{{e}^{2p}},dp}$
Problem 8 –
$latex \displaystyle \int{5{{e}^{-0.3g}},dg}$
