Use substitution, setting [latex]u=\text{−}x,[/latex] and then [latex]du=-1dx.[/latex]

Multiply the du equation by [latex]−1[/latex], so you now have [latex]\text{−}du=dx.[/latex]

Then,

[latex]\begin{array}{cc}{\displaystyle\int {e}^{\text{−}x}dx}\hfill & =\text{−}{\displaystyle\int {e}^{u}du}\hfill \\ & =\text{−}{e}^{u}+C\hfill \\ & =\text{−}{e}^{\text{−}x}+C.\hfill \end{array}[/latex]

First rewrite the problem using a rational exponent:

Using substitution, choose [latex]u=1+{e}^{x}.u=1+{e}^{x}.[/latex]

Then, [latex]du={e}^{x}dx.[/latex] We have,

Then

Here we choose to let [latex]u[/latex] equal the expression in the exponent on [latex]e[/latex].

Let [latex]u=2{x}^{3}[/latex] and [latex]du=6{x}^{2}dx..[/latex] Again, du is off by a constant multiplier; the original function contains a factor of 3[latex]x^2[/latex], not 6[latex]x^2[/latex].

Multiply both sides of the equation by [latex]\frac{1}{2}[/latex] so that the integrand in [latex]u[/latex] equals the integrand in [latex]x[/latex].

Thus,

Integrate the expression in [latex]u[/latex] and then substitute the original expression in [latex]x[/latex] back into the [latex]u[/latex] integral:

To find the price–demand equation, integrate the marginal price–demand function. First find the antiderivative, then look at the particulars.

Thus,

Using substitution, let [latex]u=-0.01x[/latex] and [latex]du=-0.01dx.[/latex] Then, divide both sides of the du equation by [latex]−0.01[/latex].

This gives,

The next step is to solve for [latex]C[/latex]. We know that when the price is [latex]$2.35[/latex] per tube, the demand is [latex]50[/latex] tubes per week. This means

Now, just solve for [latex]C[/latex]:

Thus,

If the supermarket sells [latex]100[/latex] tubes of toothpaste per week, the price would be,

The supermarket should charge [latex]$1.99[/latex] per tube if it is selling [latex]100[/latex] tubes per week.

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.6.1” here (opens in new window).

Again, substitution is the method to use. Let [latex]u=1-x,[/latex] so [latex]du=-1dx[/latex] or [latex]\text{−}du=dx.[/latex] Then [latex]\displaystyle\int {e}^{1-x}dx=\text{−}\displaystyle\int {e}^{u}du.[/latex] Next, change the limits of integration. Using the equation [latex]u=1-x,[/latex] we have

The integral then becomes

We have,

[latex]Q(t)=\int {3}^{t}dt=\frac{{3}^{t}}{\text{ln}3}+C.[/latex]

Then, at [latex]t=0[/latex] we have

[latex]Q(0)=10=\frac{1}{\text{ln}3}+C,[/latex] 

so [latex]C\approx 9.090[/latex] and we get,

[latex]Q(t)=\frac{{3}^{t}}{\text{ln}3}+9.090.[/latex]

Note: We are using [latex]10[/latex] in place of [latex]10,000[/latex] since [latex]10 ,000[/latex] bacteria are [latex]10[/latex] thousands of bacteria. We will multiple our final answer by a power of [latex]1000[/latex] at the end of our calculation to account for this. 

At time [latex]t=2,[/latex] we have,

After [latex]2[/latex] hours, there are [latex]17,282[/latex] bacteria in the dish.

This problem requires some rewriting to simplify applying the properties.

First, rewrite the exponent on [latex]e[/latex] as a power of [latex]x[/latex], then bring the [latex]x^2[/latex] in the denominator up to the numerator using a negative exponent.

We have,

Let [latex]u={x}^{-1},[/latex] the exponent on [latex]e[/latex].

Then,

Bringing the negative sign outside the integral sign, the problem now reads,

Next, change the limits of integration:

Notice that now the limits begin with the larger number, meaning we must multiply by [latex]−1[/latex]and interchange the limits.

Thus,