Question

Let \(\mathrm{f}\) and \(\mathrm{g}\) be differentiable on \([\mathrm{a}, \mathrm{b}]\) and suppose \(\mathrm{f}\) and \(\mathrm{g}^{\prime}\) are both continuous on \([\mathrm{a}, \mathrm{b}]\), then prove that\int_{a}^{b} f^{\prime}(x) g(x) d x+\int_{a}^{b} f(x) g^{\prime}(x) d x=f(b) g(b)-f(a) g(a)

Short answer

Question: Prove that for two functions f(x) and g(x), which are differentiable on the interval [a, b] and f(x) and g'(x) are continuous on the same interval, the following equation is true: $$\int_{a}^{b} f^{\prime}(x) g(x) dx + \int_{a}^{b} f(x) g^{\prime}(x) dx = f(b)g(b) - f(a)g(a)$$ Answer: We can prove this equation by using the product rule for differentiation and the fundamental theorem of calculus. Following a step-by-step process, we integrate both sides of the product rule, separate the integrals, and simplify the left-hand side to arrive at the desired result. The proof is as follows: 1. Recall the product rule for differentiation: $$(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$$ 2. Integrate both sides with respect to x over the interval [a, b]: $$\int_{a}^{b} (f(x)g(x))' dx = \int_{a}^{b} (f'(x)g(x) + f(x)g'(x)) dx$$ 3. Separate the integrals: $$\int_{a}^{b} (f(x)g(x))' dx = \int_{a}^{b} f'(x)g(x) dx + \int_{a}^{b} f(x)g'(x) dx$$ 4. Integrate the left-hand side using the fundamental theorem of calculus: $$(f(x)g(x)) |_{a}^{b} = \int_{a}^{b} f'(x)g(x) dx + \int_{a}^{b} f(x)g'(x) dx$$ 5. Simplify the left-hand side: $$f(b)g(b) - f(a)g(a) = \int_{a}^{b} f'(x)g(x) dx + \int_{a}^{b} f(x)g'(x) dx$$ Thus, the given equation is proven true.

Step-by-step solution

Recall the Product Rule for Differentiation

Recall that the product rule for differentiation states: $$(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$$

Integrate Both Sides of the Product Rule

Integrating both sides with respect to x over the interval [a, b] will give us: $$\int_{a}^{b} (f(x)g(x))' dx = \int_{a}^{b} (f'(x)g(x) + f(x)g'(x)) dx$$

Separate the Integrals

Now, separate the integrals on the right-hand side of the equation: $$\int_{a}^{b} (f(x)g(x))' dx = \int_{a}^{b} f'(x)g(x) dx + \int_{a}^{b} f(x)g'(x) dx$$

Integrate the Left-hand Side

Using the fundamental theorem of calculus, integrate the left-hand side: $$(f(x)g(x)) |_{a}^{b} = \int_{a}^{b} f'(x)g(x) dx + \int_{a}^{b} f(x)g'(x) dx$$

Simplify the Left-hand Side

Finally, simplify the left-hand side: $$f(b)g(b) - f(a)g(a) = \int_{a}^{b} f'(x)g(x) dx + \int_{a}^{b} f(x)g'(x) dx$$ And that is our desired result.

Key concepts

Product Rule for Differentiation

Understanding the product rule for differentiation is essential for anyone studying integral calculus. It's a fundamental concept that describes how to differentiate the product of two functions. When we have two functions, let's say \( f(x) \) and \( g(x) \), and we multiply them together to get \( f(x)g(x) \), the product rule helps us find the derivative of this new function.

Put simply, if you need to find the derivative of \( f(x)g(x) \), you apply the product rule as follows:
\[ (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) \]
This means you take the derivative of the first function \( f(x) \) and multiply it by the second function \( g(x) \) as it is. Then, you take the first function without changes and multiply it by the derivative of the second function \( g(x) \). Lastly, you add the two results together to get the derivative of the product.

Fundamental Theorem of Calculus

The fundamental theorem of calculus bridges the gap between differentiation and integration, which are two core operations in calculus. This theorem comes in two parts, but for the purposes of our discussion, we will focus on the part that applies to our problem.

The relevant part of the theorem states that if \( F(x) \) is an antiderivative of \( f(x) \) on an interval \( [a, b] \), then:
\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]
It tells us that the definite integral of \( f(x) \) from \( a \) to \( b \) is equal to the difference between the values of its antiderivative at \( b \) and \( a \). This is a powerful concept because it allows us to evaluate integrals by finding antiderivatives, rather than summing up infinitesimally small areas, which can be far more complex.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation, and it is used to integrate products of functions. The formula for integration by parts derives from the product rule and is given by:
\[ \int u(x) \, dv(x) = u(x)v(x) - \int v(x) \, du(x) \]
Here, \( u(x) \) and \( dv(x) \) are parts of our function where \( u(x) \) is usually chosen to be a function that simplifies when differentiated, and \( dv(x) \) is a function that is relatively easy to integrate.

The idea is to transform a complex integral into a simpler one, which is then easier to solve. This method is especially useful when faced with a product of functions such as polynomials and exponentials, logarithmic or trigonometric functions. It is important to remember that choosing \( u(x) \) and \( dv(x) \) wisely is key to simplifying the problem effectively.