Problem 68
Question
Evaluate the definite integral. \( \displaystyle \int^4_0 \frac{x}{\sqrt{1 + 2x}} \, dx \)
Step-by-Step Solution
Verified Answer
The value of the integral is \( \frac{10}{3} \).
1Step 1: Choose a Substitution
We will use substitution to simplify the integral. Let \( u = 1 + 2x \). Then, \( du = 2 \, dx \) or \( dx = \frac{1}{2} \, du \).
2Step 2: Change the Limits of Integration
Calculate the new limits of integration in terms of \( u \). When \( x = 0 \), \( u = 1 + 2(0) = 1 \). When \( x = 4 \), \( u = 1 + 2(4) = 9 \). So, the new limits are from 1 to 9.
3Step 3: Rewrite the Integral
Substitute \( x = \frac{u - 1}{2} \) and \( dx = \frac{1}{2} \, du \) into the integral. The integral becomes \[ \int_{1}^{9} \frac{\frac{u - 1}{2}}{\sqrt{u}} \cdot \frac{1}{2} \, du = \int_{1}^{9} \frac{u - 1}{4\sqrt{u}} \, du \].
4Step 4: Simplify the Integral Expression
Simplify the integral: \[ \int_{1}^{9} \left( \frac{u}{4\sqrt{u}} - \frac{1}{4\sqrt{u}} \right) du = \int_{1}^{9} \left( \frac{\sqrt{u}}{4} - \frac{1}{4\sqrt{u}} \right) du \].
5Step 5: Integrate Each Term Separately
Integrate term by term. The first term \( \int \frac{\sqrt{u}}{4} \, du \) is \( \frac{1}{4} \cdot \frac{2}{3} u^{3/2} = \frac{1}{6} u^{3/2} \). The second term \( \int \frac{1}{4\sqrt{u}} \, du \) is \( \frac{1}{4} \cdot 2 \sqrt{u} = \frac{1}{2} \sqrt{u} \).
6Step 6: Apply Limits to the Integrated Terms
Apply the integration limits to the result. This gives: \( \left[ \frac{1}{6} u^{3/2} - \frac{1}{2} \sqrt{u} \right]_{1}^{9} \).
7Step 7: Compute Definite Values
Evaluate at the upper limit: \( \frac{1}{6}(9^{3/2}) - \frac{1}{2}(9^{1/2}) = \frac{1}{6}(27) - \frac{1}{2}(3) = \frac{27}{6} - \frac{3}{2} = \frac{27}{6} - \frac{9}{6} = \frac{18}{6} = 3 \). Evaluate at the lower limit: \( \frac{1}{6}(1^{3/2}) - \frac{1}{2}(1^{1/2}) = \frac{1}{6} - \frac{1}{2} = \frac{1}{6} - \frac{3}{6} = -\frac{2}{6} = -\frac{1}{3} \).
8Step 8: Compute the Final Answer
Subtract the lower limit evaluation from the upper limit evaluation: \( 3 - \left(-\frac{1}{3}\right) = 3 + \frac{1}{3} = \frac{9}{3} + \frac{1}{3} = \frac{10}{3} \).
Key Concepts
Substitution MethodLimits of IntegrationIntegration by PartsSquare Root Function
Substitution Method
The substitution method is a powerful technique in calculus, specifically for simplifying integrals. It involves changing the variable of integration to make the integral easier to solve.
To use the substitution method, you first identify a part of the integral that can be replaced by a new variable, commonly denoted as \( u \). This helps transform the integral into a more familiar or simpler form. In our scenario, we set \( u = 1 + 2x \), which captures a complex part of the integrand. Consequently, the derivative \( du = 2 \, dx \) is used to express \( dx \) in terms of \( du \), simplifying the integral further.
Once the substitution is complete, the variable \( x \) is now expressed in terms of \( u \), leading to a transformed integral. The main advantage is that it often results in an easier integral to evaluate. Substitution is particularly useful when dealing with composite functions or integrals involving products of functions.
To use the substitution method, you first identify a part of the integral that can be replaced by a new variable, commonly denoted as \( u \). This helps transform the integral into a more familiar or simpler form. In our scenario, we set \( u = 1 + 2x \), which captures a complex part of the integrand. Consequently, the derivative \( du = 2 \, dx \) is used to express \( dx \) in terms of \( du \), simplifying the integral further.
Once the substitution is complete, the variable \( x \) is now expressed in terms of \( u \), leading to a transformed integral. The main advantage is that it often results in an easier integral to evaluate. Substitution is particularly useful when dealing with composite functions or integrals involving products of functions.
Limits of Integration
The limits of integration mark the bounds within which we're integrating. Initially, these limits are given in terms of the original variable. However, after a substitution, you need to adjust the limits to reflect the new variable.
In our example, the original limits are from \( x = 0 \) to \( x = 4 \). After substituting \( u = 1 + 2x \), the new limits need computation. When \( x = 0 \), substituting gives \( u = 1 \), and for \( x = 4 \), \( u = 9 \). Thus, our new limits for the integral are \( 1 \) to \( 9 \).
Changing limits is crucial. If ignored, it can lead to an incorrect evaluation of the integral. It ensures that the new integral accurately represents the area under the curve between the original bounds, but in the new variable context.
In our example, the original limits are from \( x = 0 \) to \( x = 4 \). After substituting \( u = 1 + 2x \), the new limits need computation. When \( x = 0 \), substituting gives \( u = 1 \), and for \( x = 4 \), \( u = 9 \). Thus, our new limits for the integral are \( 1 \) to \( 9 \).
Changing limits is crucial. If ignored, it can lead to an incorrect evaluation of the integral. It ensures that the new integral accurately represents the area under the curve between the original bounds, but in the new variable context.
Integration by Parts
Although our exercise didn't require it, understanding integration by parts is beneficial as it's another technique when substitution doesn't simplify an integral enough. It is especially handy for integrals of products of functions.
Integration by parts is derived from the product rule for differentiation and involves choosing parts of the function to differentiate and integrate. The formula is:
\[\int u \, dv = uv - \int v \, du \]where \( u \) and \( v \) are differentiable functions.
Integration by parts is derived from the product rule for differentiation and involves choosing parts of the function to differentiate and integrate. The formula is:
\[\int u \, dv = uv - \int v \, du \]where \( u \) and \( v \) are differentiable functions.
- Select \( u \) to be a part of the function that simplifies when differentiated.
- Select \( dv \) to be the remaining part.
- Then, compute \( du \) and \( v \) to use in the formula above.
Square Root Function
The square root function is a fundamental element in many calculus problems, notably when dealing with integrals. It's represented as \( \sqrt{x} \) or \( x^{1/2} \), both indicating the same operation.
In calculus, dealing with square roots inside integrals can be tricky. Substituting variables that simplify or naturally handle square roots, as done here with \( u = 1 + 2x \), is key. This can transform the square root into a simpler form, like \( u^{1/2} \), making the integral more straightforward to solve.
In calculus, dealing with square roots inside integrals can be tricky. Substituting variables that simplify or naturally handle square roots, as done here with \( u = 1 + 2x \), is key. This can transform the square root into a simpler form, like \( u^{1/2} \), making the integral more straightforward to solve.
- Understand square roots raise the number to the power of half.
- Be aware of algebraic manipulations, such as squaring both sides when solving equations involving square roots.
Other exercises in this chapter
Problem 67
Let \( \displaystyle F(x) = \int^x_2 e^{t^2} \, dt \). Find an equation of the tangent line to the curve \( y = F(x) \) at the point with \( x \)-coordinate 2.
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If \( \displaystyle f(x) = \int^{\sin x}_0 \sqrt{1 + t^2} \, dt \) and \( \displaystyle g(y) = \int^y_3 f(x) \, dx \), find \( g''(\pi/6) \).
View solution Problem 68
Which of the integrals \( \displaystyle \int^{0.5}_0 \cos (x^2) \,dx \), \( \displaystyle \int^{0.5}_0 \cos \sqrt{x} \,dx \) is larger? Why?
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