Problem 15 - University Calculus: Early Transcendentals

Question

Use the Substitution Formula in Theorem 7 to evaluate the integrals in Exercises \(1-46\). $$\int_{0}^{1} \sqrt{t^{5}+2 t}\left(5 t^{4}+2\right) d t$$

Step-by-Step Solution

Answer

The integral evaluates to \( 2\sqrt{3} \).

Step 1: Identify the Inner Function and Its Derivative

The integral given is \( \int_{0}^{1} \sqrt{t^{5}+2t}(5t^4+2) \, dt \). The substitution formula is usually applied where we identify an inner function \( u(t) \), such that the integral simplifies. In this case, let \( u = t^{5} + 2t \).

Step 2: Differentiate the Inner Function

Next, differentiate the function \( u \) with respect to \( t \): \[ du = (5t^4 + 2) \, dt \]. Notice this matches the remaining function in the integral. Thus, this substitution will simplify our integration.

Step 3: Rewrite the Integral in Terms of \( u \)

Using the substitution \( u = t^5 + 2t \) and \( du = (5t^4 + 2) \, dt \), rewrite the integral as \( \int \sqrt{u} \, du \).

Step 4: Change the Limits of Integration

Substituting the limits of integration requires us to find \( u \) when \( t = 0 \) and \( t = 1 \):- At \( t = 0 \), \( u = 0^5 + 2(0) = 0 \).- At \( t = 1 \), \( u = 1^5 + 2(1) = 3 \).Thus, the limits become \( 0 \) and \( 3 \).

Step 5: Evaluate the Integral

Now integrate \( \int_{0}^{3} \sqrt{u} \, du \). This integral can be solved by using the power rule of integration: \( \int u^{n} \, du = \frac{u^{n+1}}{n+1} + C \). Here, \( n = \frac{1}{2} \), so the integral becomes:\[ \int \sqrt{u} \, du = \int u^{\frac{1}{2}} \, du = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3} u^{\frac{3}{2}} + C \].

Step 6: Substitute Back the Limits

Finally, apply the new limits to evaluate the definite integral: \[ \frac{2}{3} u^{\frac{3}{2}} \bigg|_{0}^{3} = \frac{2}{3} (3^{\frac{3}{2}}) - \frac{2}{3} (0^{\frac{3}{2}}) = \frac{2}{3} \times 3\sqrt{3} = 2\sqrt{3} \]

Key Concepts

Integration by substitution

The method of integration by substitution is a powerful tool in calculus. It simplifies integrals by changing the variable of integration to a new variable. This technique is akin to reversing the chain rule for derivatives.
Imagine having a complex function inside the integral that is difficult to integrate directly.

• First, identify an inner function, often denoted as \( u \), which makes the problem look less complex.
• Next, find the derivative of this function with respect to the original variable, in our case, \( t \). This gives you \( du \).
• Once \( u \) and \( du \) are known, rewrite the integral in terms of the new variable \( u \).

For example, in the exercise, the substitution \( u = t^5 + 2t \) greatly simplifies the integral, transforming it into \( \int \sqrt{u} \, du \).This results in an integral that can be more easily evaluated using basic integration techniques.
The new limit of integration needs to be defined in terms of \( u \), which often involves evaluating the substitution at the original limits.

Definite integrals

Definite integrals are a basic concept in calculus, representing the accumulation of quantities and often used to find areas under curves.
Unlike indefinite integrals, definite integrals have limits of integration that define the interval over which the function is integrated.

• In the substitution method, once you have solved the integral in terms of \( u \), you need to change the limits of integration to match the new variable.
• This involves substituting the original limits into the expression for \( u \). For example, when \( t = 0 \), \( u = 0 \) and similarly for \( t = 1 \), \( u = 3 \). Thus, the integral is now definite from \( 0 \) to \( 3 \).

The evaluated integral gives a numerical result, illustrating the net accumulated quantity over the interval. In our exercise, solving \( \int_{0}^{3} \sqrt{u} \, du \) yields the result \( 2\sqrt{3} \).
This value represents the accumulated area under the curve \( \sqrt{u} \) from \( u = 0 \) to \( u = 3 \).

Calculus problem-solving

Problem-solving in calculus often requires a strategic approach, combining multiple rules and techniques based on the problem's requirements.
When faced with complex integrals, deciding the most suitable method is crucial.

• In many cases, integration by substitution emerges as an optimal choice, especially when a part of the integrand and its derivative are present in the integral.
• Once the method is chosen, carefully follow the steps, ensuring correctness in substitution and integration limits.

This structured approach simplifies the problem-solving process and helps in obtaining accurate results.
For this exercise, using integration by substitution transformed a tricky expression into a more manageable one.
Finding the right substitution and adjusting the limits enabled easy application of the power rule to evaluate the definite integral.