Problem 302
Question
In the following exercises, evaluate the indefinite integral \(\int f(x) d x\) with constant \(C=0\) using u-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of \(C\) that would need to be added to the antiderivative to make it equal to the definite integral \(F(x)=\int_{a}^{x} f(t) d t,\) with a the left endpoint of the given interval. $$ \int(x+2) e^{-x^{2}-4 x+3} d x \text { over }[-5,1] $$
Step-by-Step Solution
Verified Answer
Integrate using u-substitution:
\( \int (x+2)e^{-x^2 - 4x + 3} \approx -\frac{1}{2}e^{-x^2 - 4x + 3}\), and graph over
\([-5,1]\). No additional C needed.
1Step 1: Identify the Inner Function for U-Substitution
In integration with u-substitution, we start by identifying the part of the function that we want to substitute with a new variable. For the integrand \( (x+2) e^{-x^{2}-4x+3} \),notice that the expression \( -x^2 - 4x + 3 \) inside the exponent is a suitable choice for substitution. Let \( u = -x^2 - 4x + 3 \).
2Step 2: Calculate the Differential du
We need to find \( du \) based on our substitution. Calculate the derivative of \( u \):\[ \frac{du}{dx} = -2x - 4 \].This gives us \( du = (-2x - 4) dx \) or \(-1(2x + 4)dx = du\).
3Step 3: Simplify the Differential
To express the integrand in terms of \( u \), we need to match the differential. Notice that \(-(2x + 4) = -(2(x + 2))\). This means we have \(-1(x + 2)dx = \frac{1}{2}du\), using this substitute in the integral.
4Step 4: Substitute into the Integral
Replace the identified parts with \( u \) and the simplified differential:\[ \int (x+2)e^{-x^2 - 4x + 3} dx = \int -\frac{1}{2} e^{u} du \]. This simplifies the integral expression.
5Step 5: Integrate with respect to u
The integral of \( -\frac{1}{2}e^{u} \) is \( -\frac{1}{2} \int e^{u} du \). The integral of \( e^{u} \) is \( e^{u} \), so the solution to the integral is:\[ -\frac{1}{2} e^{u} + C \].
6Step 6: Substitute Back the original variable
Replace \( u \) with the original expression in terms of \( x \):\[ -\frac{1}{2}e^{-x^2 - 4x + 3} + C \].
7Step 7: Graph the Function and Its Antiderivative
Graph the function \( (x+2)e^{-x^2 - 4x + 3} \) and the antiderivative \( -\frac{1}{2}e^{-x^2 - 4x + 3} \) over the interval \([-5, 1]\). The antiderivative will likely show a smooth curve while the function's graph indicates its mathematical behavior.
8Step 8: Estimate the Value of C for the Definite Integral
Consider the definite integral \( F(x)=\int_{-5}^{x} (x+2)e^{-x^2 - 4x + 3} dt \). Adjust the constant \( C \) such that the indefinite integral evaluated agrees with the limits of the definite integral as the starting point matches the left endpoint of the interval \([-5,1]\). As the problem suggests \( C=0\) will fit the context well since the integral calculation assumes the specific offset.
Key Concepts
Indefinite IntegralAntiderivativeDefinite IntegralGraphing Functions
Indefinite Integral
An indefinite integral is a type of integral that represents not just one, but a whole family of functions. These functions are known as antiderivatives. When you calculate an indefinite integral, you find a function whose derivative is equal to the original function being integrated. Indefinite integrals do not have specific bounds or limits. This means that when we solve an indefinite integral, we add a constant, traditionally called "C," to the solution.
The reason behind adding the constant is that differentiating a constant gives zero, and hence there are infinitely many functions that could satisfy the derivative condition — each differing by some constant. The notation we typically use is \(\int f(x) \, dx = F(x) + C\). The solved example demonstrates this concept by integrating the function\((x+2) e^{-x^2-4x+3} \), resulting in the expression\(-\frac{1}{2}e^{-x^2-4x+3} + C\). Here, "C" accounts for any constant value that could be included in the antiderivative.
The reason behind adding the constant is that differentiating a constant gives zero, and hence there are infinitely many functions that could satisfy the derivative condition — each differing by some constant. The notation we typically use is \(\int f(x) \, dx = F(x) + C\). The solved example demonstrates this concept by integrating the function\((x+2) e^{-x^2-4x+3} \), resulting in the expression\(-\frac{1}{2}e^{-x^2-4x+3} + C\). Here, "C" accounts for any constant value that could be included in the antiderivative.
Antiderivative
An antiderivative of a function is essentially the reverse operation of differentiation. If you have a function, your goal in finding the antiderivative is to identify another function whose derivative matches the initial function. In everyday terms, if you know how something changes, the antiderivative tells you what the original function looks like.
Take for example the function used earlier \((x+2) e^{-x^2-4x+3} \). By applying the u-substitution method, we find that its antiderivative is\(-\frac{1}{2} e^{-x^2-4x+3} + C\). The antiderivative provides a broader view of how the quantity evolves; you would effectively 'retrace your steps' to determine this bigger picture.
Understanding antiderivatives is crucial because they are foundational building blocks for both definite and indefinite integrals, bridging gaps between calculus and real-world applications.
Take for example the function used earlier \((x+2) e^{-x^2-4x+3} \). By applying the u-substitution method, we find that its antiderivative is\(-\frac{1}{2} e^{-x^2-4x+3} + C\). The antiderivative provides a broader view of how the quantity evolves; you would effectively 'retrace your steps' to determine this bigger picture.
Understanding antiderivatives is crucial because they are foundational building blocks for both definite and indefinite integrals, bridging gaps between calculus and real-world applications.
Definite Integral
Definite integrals are slightly different from indefinite ones because they involve limits of integration and yield a numeric value rather than a family of functions. They represent the total accumulation or "net" change over a specific interval \([a, b]\). The definite integral takes into consideration the start and endpoint, giving an exact area under the curve between these two points.
The problem involved finding a value for "C" such that the antiderivative matches the definite integral when computed over\([-5, 1]\). By setting\(C = 0\), we align the indefinite integral with the calculation performed by the definite integral starting at the leftmost endpoint of the interval. This helps to ensure that both values reconcile over the given range, demonstrating how constants in indefinite integrals adjust to match the confines of a definite integral.
Evaluating definite integrals is especially significant in physics and engineering when calculating things like total displacement, area, or work done over a period.
The problem involved finding a value for "C" such that the antiderivative matches the definite integral when computed over\([-5, 1]\). By setting\(C = 0\), we align the indefinite integral with the calculation performed by the definite integral starting at the leftmost endpoint of the interval. This helps to ensure that both values reconcile over the given range, demonstrating how constants in indefinite integrals adjust to match the confines of a definite integral.
Evaluating definite integrals is especially significant in physics and engineering when calculating things like total displacement, area, or work done over a period.
Graphing Functions
Graphing is an essential aspect that visualizes functions and antiderivatives. Graphing the initial function and its antiderivative helps compare their behaviors and understand their interactions over a given interval. In this exercise, you would graph the function \((x+2)e^{-x^2-4x+3} \) alongside its antiderivative \(-\frac{1}{2} e^{-x^2-4x+3} \) over the range \([-5, 1]\).
The process of plotting both functions might reveal features like peaks and valleys or pedestrian trends that are not evident from equations alone. The graph of the antiderivative often appears smoother and less erratic than that of the original function since it encompasses a broader perspective of the mathematical trend.
The process of plotting both functions might reveal features like peaks and valleys or pedestrian trends that are not evident from equations alone. The graph of the antiderivative often appears smoother and less erratic than that of the original function since it encompasses a broader perspective of the mathematical trend.
- Graphs help verify calculations by providing a visual representation.
- You can identify discrepancies or errors in your calculations by closely examining trends on the graph.
- Graphing functions and their antiderivatives over specified intervals allows for direct comparison to ensure consistency between mathematical calculations and graphical interpretations.
Other exercises in this chapter
Problem 300
In the following exercises, evaluate the indefinite integral \(\int f(x) d x\) with constant \(C=0\) using \(u\) -substitution. Then, graph the function and the
View solution Problem 301
In the following exercises, evaluate the indefinite integral \(\int f(x) d x\) with constant \(C=0\) using u-substitution. Then, graph the function and the anti
View solution Problem 303
In the following exercises, evaluate the indefinite integral \(\int f(x) d x\) with constant \(C=0\) using u-substitution. Then, graph the function and the anti
View solution Problem 304
In the following exercises, evaluate the indefinite integral \(\int f(x) d x\) with constant \(C=0\) using u-substitution. Then, graph the function and the anti
View solution