Question

Determine whether the following statements are true and give an explanation or counterexample. Assume \(f, f^{\prime},\) and \(f^{\prime \prime}\) are continuous functions for all real numbers. a. \(\int f(x) f^{\prime}(x) d x=\frac{1}{2}(f(x))^{2}+C\) b. \(\int(f(x))^{n} f^{\prime}(x) d x=\frac{1}{n+1}(f(x))^{n+1}+C, n \neq-1\) c. \(\int \sin 2 x \, d x=2 \int \sin x \, d x\) d. \(\int\left(x^{2}+1\right)^{9} d x=\frac{\left(x^{2}+1\right)^{10}}{10}+C\) e. \(\int_{a}^{b} f^{\prime}(x) f^{\prime \prime}(x) d x=f^{\prime}(b)-f^{\prime}(a)\)

Short answer

Question: Determine if the given statements are true or false and provide an explanation or counterexample. a. True or False: \(\int f(x)f'(x)dx = \frac{1}{2}(f(x))^2 + C\) Answer: True b. True or False: \(\int(f(x))^{n}f'(x)dx=\frac{1}{n+1}(f(x))^{n+1}+C, n \neq -1\) Answer: Conditionally True for \(n=1\) but False in general. c. True or False: \(\int \sin 2x \, dx=2 \int \sin x \, dx\) Answer: False d. True or False: \(\int(x^2+1)^9dx = \frac{(x^2+1)^{10}}{10} + C\) Answer: False e. True or False: \(\int_a^b f''(x)f'(x)dx = f'(b) - f'(a)\) Answer: True

Step-by-step solution

Use Integration by Parts

To solve this integral, we can apply integration by parts, which states that \(\int u \, dv = uv - \int v \, du\). Let \(u = f(x)\) and \(dv = f'(x)dx\). Then we get \(du = f'(x)dx\) and \(v = f(x)\). So we have: $$\int f(x)f'(x)dx = f(x)f(x) - \int f(x)f'(x)dx$$

Verify the Equation

Now we have the equation \(\int f(x)f'(x)dx = (f(x))^2 - \int f(x)f'(x)dx\). Adding the integral term to both sides, we get: $$2\int f(x)f'(x)dx = (f(x))^2$$ Dividing by 2, we get the result, $$\int f(x)f'(x)dx = \frac{1}{2}(f(x))^2+C$$ Hence, the statement is True. #b. True or False: \(\int(f(x))^{n}f'(x)dx=\frac{1}{n+1}(f(x))^{n+1}+C, n \neq -1\)

Use Integration by Parts

We'll again apply integration by parts. Let \(u = (f(x))^n\) and \(dv = f'(x)dx\). Then we get \(du = nf(x)^{n-1}f'(x)dx\) and \(v = f(x)\). So we have: $$\int(f(x))^n f'(x)dx = (f(x))^{n+1} - \int nf(x)^{n-1}(f(x))^2 f'(x)dx$$

Verify the Equation

We can't simplify any further, and it doesn't match the given result. This doesn't confirm that the statement is false, but we can give an easy counterexample.

Counterexample with \(n=1\)

Let's consider the case when \(n=1\). Then we have: $$\int f(x)f'(x)dx \stackrel{?}{=} \frac{1}{1+1}(f(x))^{1+1}+C = \frac{1}{2}(f(x))^2+C$$ We have already proved in part (a) that his equation is true. So, in this case, the statement is correct; however, this does not guarantee it being true for all other \(n\). In conclusion, we can't claim the statement is universally true or false, so we say it is conditionally True for \(n=1\) but False in general since we cannot prove the result. #c. True or False: \(\int \sin 2x \, dx=2 \int \sin x \, dx\)

Integration of \(\sin2x\)

To find the integral of \(\sin 2x\), notice the chain rule derivative in the function and perform substitution. Let \(u = 2x, du = 2dx\). Then: $$\int \sin (2x)dx = \frac{1}{2} \int \sin(u)du$$ Now, we know that the integral of \(\sin(u)\) is \(-\cos(u)\). So we obtain, $$\int \sin (2x)dx = -\frac{1}{2}\cos(2x) + C$$

Compare the Functions

The given statement claims that \(\int \sin 2x \, dx=2 \int \sin x \, dx\). We have found that the left side equals \(-\frac{1}{2}\cos(2x) + C\). Since these expressions are not equivalent, the statement is False. #d. True or False: \(\int(x^2+1)^9dx = \frac{(x^2+1)^{10}}{10} + C\)

Integration

In this case, we will perform a simple substitution. Let \(u = x^2 + 1\) and therefore, \(du = 2x dx\). We have: $$\int(x^2+1)^9dx = \frac{1}{2} \int u^9 du$$ Integrating, we get: $$\int(x^2+1)^9dx = \frac{1}{2}\cdot\frac{u^{10}}{10}+C = \frac{(x^2+1)^{10}}{20} + C$$

Compare the Functions

We have obtained the result: \(\int(x^2+1)^9dx=\frac{(x^2+1)^{10}}{20}+C\), while the given statement claims that \(\int(x^2+1)^9dx=\frac{(x^2+1)^{10}}{10}+C\). Since these expressions are not the same, the statement is False. #e. True or False: \(\int_a^b f''(x)f'(x)dx = f'(b) - f'(a)\)

Use the Fundamental Theorem of Calculus

This question tests our understanding of the second part of the Fundamental Theorem of Calculus. We have: $$\int_a^b f''(x)f'(x)dx = \int_a^b \frac{d}{dx}(f'(x))f'(x)dx$$ Using the properties of the integral and differentials, we can convert the above expression into: $$\int_a^b \frac{d}{dx}(f'(x))f'(x)dx = [f'(x)]_{a}^{b} = f'(b) - f'(a)$$ Thus, this statement is True.

Key concepts

Calculus

Calculus is a branch of mathematics centered on the study of how things change. It allows us to calculate the rate at which quantities evolve over time or space, such as velocity in physics or the slope of a curve in geometry. Fundamental to calculus are the concepts of differentiation and integration, which are two sides of the same coin. Differentiation gives us the rate of change (or the derivative), while integration sums (or 'integrates') these rates over time to find out how much things have changed. An easy analogy would be comparing differentiation to figuring out your car's speed at any given moment, while integration would tell you how far you've traveled after some time, using the speed at every moment along the way.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is pivotal in understanding the relationship between differentiation and integration. It essentially bridges the gap between the two, and can be thought of in two parts:

• The first part tells us that if we have a continuous function, the integral of its derivative over an interval is the same as the difference in the function's values at the end points of that interval. A practical understanding of this is if you're watching a marathon and you note down the runners' positions at two points in time; the integral of their speed (the derivative of their position) would give you the distance they've covered between those times.
• The second part emphasizes that any continuous function has an antiderivative, and the definite integral of a function over an interval can be calculated using any of its antiderivatives. This can be likened to using the known cumulative distance traveled over time to deduce the trips' start and end positions.

This theorem is essential when evaluating definite and indefinite integrals, making it a cornerstone of our calculus toolkit.

Indefinite Integral

When it comes to integrals, there are two main types: indefinite and definite. An indefinite integral, often referred to as an antiderivative, represents a family of functions that share the same derivative. By integrating a function without specific limits, we're essentially finding the general formula for the area under its curve. For example, the indefinite integral of a function like f(x) = 2x is written as t t f(x) dx = x^2 + C, where C represents an arbitrary constant because when we differentiate x^2 + C, the constant vanishes and we're left with 2x, the original function. So, the indefinite integral gives us a set of possible functions all differing by a constant whose graphs all have the same slope at any given point.

Definite Integral

In contrast to the indefinite integral, the definite integral has boundaries. It computes the net area under the curve of a function between two specified points. This calculation tells us the accumulated total from the start point to the end point, factoring in both the highs and lows along the way. Going back to our marathon analogy from the Fundamental Theorem of Calculus, if you want to know precisely how far a runner has run between the 5-mile mark and the 10-mile mark, you would use a definite integral.Mathematically, the definite integral from a to b of the function f(x) is denoted as t_a^b f(x) dx, and according to the Fundamental Theorem of Calculus, it can be computed using an antiderivative, F(x), of the function—t_a^b f(x) dx = F(b) - F(a). Here, we subtract the value of the antiderivative at a from its value at b, giving us the net 'area'—which could represent distance, total cost, or numerous other cumulative quantities, depending on the scenario.