Problem 49

Question

, use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{0}^{1} \cos (3 x-3) d x $$

Step-by-Step Solution

Verified

Answer

The definite integral evaluates to $ \frac{\sin(3)}{3} $.

1Step 1: Identify the Substitution

For the given integral $ \int_{0}^{1} \cos (3x-3) \, dx $, we identify the substitution $ u = 3x - 3 $. This substitution simplifies the expression inside the cosine function.

2Step 2: Calculate the Derivative and Solve for dx

Differentiate the substitution $ u = 3x - 3 $ to find $ \frac{du}{dx} = 3 $. Solving for $ dx $, we get $ dx = \frac{du}{3} $.

3Step 3: Change the Limits of Integration

Substitute the limits of $ x $ into the expression for $ u $. When $ x = 0 $, $ u = 3(0) - 3 = -3 $. When $ x = 1 $, $ u = 3(1) - 3 = 0 $. The new limits for $ u $ are from $-3$ to $0$.

4Step 4: Substitute in the Integral

Substitute $ u = 3x - 3 $ and $ dx = \frac{du}{3} $ into the integral, and adjust the limits of integration: \[ \int_{-3}^{0} \cos(u) \cdot \frac{1}{3} \, du \].

5Step 5: Factor Out the Constant

Factor the constant $ \frac{1}{3} $ out of the integral: \[ \frac{1}{3} \int_{-3}^{0} \cos(u) \, du \].

6Step 6: Integrate the Function

Integrate $ \cos(u) $ with respect to $ u $: $ \int \cos(u) \, du = \sin(u) $.

7Step 7: Evaluate the Definite Integral

Evaluate $ \frac{1}{3} [\sin(u)]_{-3}^{0} = \frac{1}{3} (\sin(0) - \sin(-3)) $. Since $ \sin(0) = 0 $ and $ \sin(-3) = -\sin(3) $, the result is $ \frac{1}{3} (0 - (-\sin(3)) = \frac{\sin(3)}{3} $.

Key Concepts

Integration by SubstitutionCalculus Problem SolvingDefinite IntegralsTrigonometric Integrals

Integration by Substitution

Integration by Substitution is akin to solving a puzzle piece by piece. The basic idea is to replace a challenging function with a simpler one, often by utilizing a substitution variable. In this particular problem, the integral $ \int_{0}^{1} \, \cos (3x-3) \, dx $ seems tough due to the expression inside the cosine function.

Firstly, identify a substitution. Here, $ u = 3x - 3 $ simplifies the integrand.
This substitution turns the complex problem into a more manageable form.

Apply differentiation: find $ \frac{du}{dx} = 3 $, leading to $ dx = \frac{du}{3} $. This is crucial as it helps us replace $ dx $ in the integral with respect to our new variable $ u $.
This technique transforms our original integral into one that is easier to evaluate.

Calculus Problem Solving

Approaching calculus problems efficiently is all about strategy.

Start by completely understanding the question.
Next, decide on the right method - here, substitution is chosen to handle the given integral.
Every step involves simplifying and transforming the problem.

Once you substitute $ u = 3x - 3 $ and solve for $ dx $, ensure to adjust the limits of integration. When $ x $ changes, $ u $ must reflect those changes. Therefore, switch the calculation from the original limits $ 0 $ to $ 1 $ for $ x $ to the new ones, $ -3 $ to $ 0 $ for $ u $.
Every detail, even simple arithmetic changes and constants like $ \frac{1}{3} $, is significant for accurate problem-solving.

Definite Integrals

Definite integrals allow us to calculate the net area under a curve over a specific interval. Unlike indefinite integrals, definite integrals require limits of integration.

These limits define the start and end points on the curve.
The substitution technique necessitates changing these limits to correspond to the new variable.

In this instance, we're moving from $ x $-based limits $ 0 $ to $ 1 $ to $ u $-based limits $ -3 $ to $ 0 $. With these new limits, integrating $ \cos(u) $ results in $ \sin(u) $, leading us to calculate the difference $ \sin(0) - \sin(-3) $.
This approach shows the balance between transformation of the function and limits to acquire a simpler path to the solution.

Trigonometric Integrals

Dealing with trigonometric integrals, such as $ \cos(u) $, might seem complex, but they follow straightforward rules. The major part is identifying the integral form.

Transform $ \int \cos(u) \, du $ to $ \sin(u) $, a basic anti-derivative relationship.
This step bridges the substitution process and the final evaluation of the integral.

Conclude by evaluating $ \sin(u) $ within the new limits $ -3 $ to $ 0 $. After substituting back and calculating $ \frac{1}{3} [\sin(0) - \sin(-3)] $, we leverage trigonometric properties: $ \sin(0) = 0 $ and $ \sin(-3) = -\sin(3) $.
This ultimately yields $ \frac{\sin(3)}{3} $ as the result, showcasing the power of substitution and trigonometric rules.

Problem 49

Problem 50

Other exercises in this chapter

Problem 49

If $f$ is periodic with period $p$, then $$ \int_{a}^{a+p} f(x) d x=\int_{0}^{p} f(x) d x $$ Convince yourself that this is true by drawing a picture and th

View solution

Problem 49

Find $\lim _{x \rightarrow 0} \frac{1}{x} \int_{0}^{x} \frac{1+t}{2+t} d t$.

View solution

Problem 50

Sketch the graph of the given function over the interval $[a, b] ;$ then divide $[a, b]$ into $n$ equal subintervals. Finally, calculate the area of the c

View solution

Problem 50

Find $\lim _{x \rightarrow 1} \frac{1}{x-1} \int_{1}^{x} \frac{1+t}{2+t} d t$.

View solution