Question

Integrating derivatives Use integration by parts to show that if \(f^{\prime}\) is continuous on \([a, b],\) then $$\int_{a}^{b} f(x) f^{\prime}(x) d x=\frac{1}{2}\left(f(b)^{2}-f(a)^{2}\right)$$.

Short answer

Question: Show that if the derivative of the function \(f\) is continuous on the interval \([a, b]\), then the integral of the product \(f(x)f'(x)\) can be represented as \(\frac{1}{2}(f(b)^2 - f(a)^2)\). Answer: We used integration by parts to show that \(\int_{a}^{b} f(x)f'(x)dx = \frac{1}{2}(f(b)^2 - f(a)^2)\).

Step-by-step solution

Integration by parts formula

Recall the integration by parts formula: $$\int u(x) v'(x) dx = u(x)v(x) - \int u'(x) v(x) dx$$ We will need to select appropriate functions \(u(x)\) and \(v'(x)\) for this problem.

Choose \(u(x)\) and \(v'(x)\)

Let's choose the functions as follows: \(u(x) = f(x)\) and \(v'(x) = f'(x)\). We'll then need to compute \(u'(x)\) and find \(v(x)\).

Compute \(u'(x)\) and find \(v(x)\)

Since we have chosen \(u(x) = f(x)\), its derivative, \(u'(x)\), is simply \(f'(x)\). To find \(v(x)\), we need to integrate \(v'(x)\): $$v(x) = \int f'(x) dx = f(x) + C$$ where \(C\) is the constant of integration. However, since we will be doing a definite integral, the constant \(C\) will cancel out and can be ignored.

Apply the integration by parts formula

Now, we can apply the integration by parts formula using \(u(x) = f(x)\), \(u'(x) = f'(x)\), and \(v(x) = f(x)\): $$\int_{a}^{b} f(x) f'(x) dx = \left[f(x)^2\right]_a^b - \int_{a}^{b} f'(x) f(x) dx$$

Simplify the integral and compute the result

We will now simplify the integral and compute the result: $$\int_{a}^{b} f(x) f'(x) dx = \left[f(b)^2 - f(a)^2\right] - \int_{a}^{b} f'(x) f(x) dx$$ Now, let's move the second integral to the other side of the equation: $$\int_{a}^{b} f(x) f'(x) dx + \int_{a}^{b} f'(x) f(x) dx = f(b)^2 - f(a)^2$$ Since the integrals are integrating the same function over the same interval, they must be equal. Thus, we can rewrite the equation as: $$2\int_{a}^{b} f(x) f'(x) dx = f(b)^2 - f(a)^2$$ Finally, we can isolate the integral we want on one side of the equation to obtain the desired result: $$\int_{a}^{b} f(x) f'(x) dx = \frac{1}{2}(f(b)^2 - f(a)^2)$$

Key concepts

Definite Integrals

A definite integral is a powerful tool in calculus used to calculate the area under a curve on a specified interval. It is represented as \( \int_{a}^{b} f(x) \,dx \), where \( f(x) \) is the function we're integrating, and \( [a,b] \) are the limits of integration, defining the interval on the x-axis over which we are calculating the area.

When dealing with definite integrals, it's crucial to understand that we are summing an infinite number of infinitesimally small rectangles under the curve, giving us the total 'accumulated' area from \( a \) to \( b \). The process of finding this area often involves finding an antiderivative - a function \( F(x) \) such that \( F'(x) = f(x) \) - and evaluating it at the upper and lower limits of integration.

Derivatives

In calculus, the derivative of a function represents the rate at which the function's value changes with respect to a change in its input value. In mathematical terms, if you have a function \( f(x) \), its derivative \( f'(x) \) is the limit of the average rate of change as the interval becomes infinitesimally small. It's the slope of the tangent line to the function at any point.

If \( f(x) \) describes a position over time, then its derivative \( f'(x) \) gives us the velocity; the derivative of velocity with respect to time gives us acceleration. This concept is fundamental for solving many problems in physics, engineering, and beyond. Derivatives are not only about rates of change; they also play an essential role in finding local maxima and minima of functions, allowing us to understand the behavior of the functions in greater detail.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, two cornerstone operations in calculus. It is divided into two parts: the First Part establishes the relationship between differentiation and integration, ensuring that if a function is continuous over an interval, then the integral of a function from \( a \) to \( b \) is the antiderivative evaluated at \( b \) minus the antiderivative evaluated at \( a \) - in other words, \( \int_{a}^{b} f(x) \,dx = F(b) - F(a) \) where \( F'(x) = f(x) \).

The Second Part tells us that if we have a function that is defined as an integral of another function, then the derivative of this function gives us the original function back. It affirms the mirrored relationship between the derivative and the integral, providing a method to evaluate definite integrals without the need for computing limits of Riemann sums directly.