Question

In Exercises \(53-66,\) make a \(u\) -substitution and integrate from \(u(a)\) to \(u(b) .\) $$\int_{0}^{3} \sqrt{y+1} d y$$

Short answer

The result of the integral is \(\frac{14}{3}\).

Step-by-step solution

Make the u-substitution

Let's start with making the substitution \(u=y+1\). This should simplify the function into a form that is easier to integrate. Remember to also calculate the differential \(du\). It will be \(du=dy\).

Adjust the limits of integration

After this substitution, adjust the limits of integration. Initially, when \(y=0\), we now have \(u=0+1=1\). And when \(y=3\), we get \(u=3+1=4\), where \(u\) is our limits or bounds for \(u\).

Replace in the integral and integrate

Now we replace all variables in the integral. So, from \(\int_{0}^{3} \sqrt{y+1}\, dy\) it now becomes \(\int_{1}^{4} \sqrt{u}\, du\). Now, we can easily calculate this integral. The integral of \(\sqrt{u}\) is \(\frac{2}{3}u^{3/2}\). Applying the limits we get \[\frac{2}{3}u^{3/2}\bigg|_{1}^{4} = \frac{2}{3}*(4^{3/2}) - \frac{2}{3}*(1^{3/2})\]

Simplify the result

Finally, we can simplify those expressions to get the final result. So the result will be \(\frac{2}{3}*8 - \frac{2}{3}*1 = \frac{16}{3} - \frac{2}{3} = \frac{14}{3}\).

Key concepts

Definite Integral

A definite integral is a fundamental concept in calculus, which provides a way to calculate the accumulated total of a function over a certain interval. The concept of the definite integral can be thought of as the 'net area' under a curve from one point to another.
To calculate a definite integral, you need to identify two main elements: the function to be integrated (known as the integrand) and the limits of integration (the two points that define the interval over which we integrate). The notation \(\int_{a}^{b} f(x)\, dx\) designates the integral of the function f(x) with respect to x, from the lower limit a to the upper limit b.
For instance, in our exercise, the definite integral is expressed as \(\int_{0}^{3} \sqrt{y+1}\, dy\), where the function inside the integral sign is \(\sqrt{y+1}\), and the interval is from y = 0 to y = 3. When calculated, this definite integral represents the total 'area' under the curve of \(\sqrt{y+1}\) between these limits.

Integration by Substitution

Integration by substitution is a method used to simplify an integral by changing the variable of integration to make the function easier to integrate. It is sometimes called 'u-substitution.' This technique is particularly useful when dealing with functions that are compositions of other functions.

Steps for Performing U-Substitution:

• Select a substitution that simplifies the integrand. Often, we try to find a function inside the integrand whose derivative is also present.
• Determine the differential of the substitution. This means calculating du in terms of dx (or whatever variable you are using).
• Rewrite the integral in terms of the new variable u and du.
• Perform the integration with respect to u.
• Substitute back to the original variable if necessary or directly compute the result if the new limits in terms of u are used.

In the given problem, the substitution is \(u = y + 1\), because it simplifies the integrand \(\sqrt{y+1}\) to \(\sqrt{u}\), and the differential is straightforward, \(du = dy\).

Integral Calculation

The process of integral calculation involves finding the antiderivative (also called the indefinite integral) of a function and then using the limits of integration to evaluate the result. The antiderivative is essentially the reverse process of differentiation.
During this calculation, you apply fundamental integration rules and techniques, such as the power rule or integration by parts, to find the function that, when differentiated, yields the original integrand. Once the antiderivative is found, you use the limits from the definite integral to extract the final value.
In our example, after applying u-substitution, we have the integral \(\int_{1}^{4} \sqrt{u}\, du\), which is easier to handle. Here, the integral of \(\sqrt{u}\) is \(\frac{2}{3}u^{\frac{3}{2}}\), by using the power rule for integration. Applying definite integral limits to this antiderivative yields the exact value of the total area under the original function within the specified interval.

Changing the Limits of Integration

After performing a u-substitution in a definite integral, we must also change the limits of integration to match the new variable of integration. This ensures we're calculating the integral over the same region, just with our new, simpler variable.
To do this, we:

• Plug the original lower limit of integration into our substitution equation to find the new lower limit in terms of u.
• Repeat this process with the original upper limit to find the new upper limit in terms of u.

These new limits replace the original ones, and we integrate within this new interval. Changing the limits is crucial because it allows us to evaluate the integral without reverting back to the original variable, thereby simplifying the calculation.
In the provided exercise, the initial limits were 0 and 3 for y. After the substitution \(u = y + 1\) was made, these limits changed to 1 and 4 for u. Integrating with respect to u and applying these new limits instantly gives us the result of the definite integral.