Question

Derive the formula $$ \int_{0}^{x}\left(\int_{0}^{t} f(z) d z\right) d t=\int_{0}^{x} f(t)(x-t) d t $$ using integration by parts.

Short answer

Both integrals are equivalent, verified through integration by parts.

Step-by-step solution

Understand the Problem

We need to show that the double integral \( \int_{0}^{x}\left(\int_{0}^{t} f(z) d z\right) d t \) can be rewritten as \( \int_{0}^{x} f(t)(x-t) d t \) using integration by parts.

Recognize the Inner Integral

The expression \( \int_{0}^{t} f(z) dz \) is an antiderivative evaluated at \( z = t \) and \( z = 0 \). This means we're integrating \( f(z) \) from \( 0 \) to \( t \).

Differentiate Under the Integral Sign

Using the Fundamental Theorem of Calculus, the derivative of \( \int_{0}^{t} f(z) dz \) with respect to \( t \) is \( f(t) \). This implies that the outer integral \( \int_{0}^{x} \left( \int_{0}^{t} f(z) dz \right) dt \) needs an adjustment for integration by parts.

Set Up Integration by Parts

Integration by parts is given by \( \int u \, dv = uv - \int v \, du \). Let \( u = \int_{0}^{t} f(z) dz \) and \( dv = dt \). This makes \( du = f(t) dt \) (by differentiating \( u \)), and \( v = t \) when integrating \( dv = dt \).

Apply Integration by Parts

According to integration by parts, \( \int_{0}^{x} u \, dv = [uv]_{0}^{x} - \int_{0}^{x} v \, du \). Substituting, we get: \[ \left[ t \int_{0}^{t} f(z) dz \right]_{0}^{x} - \int_{0}^{x} t \, f(t) \, dt. \]

Evaluate the Boundary Terms

Evaluate \( \left[ t \int_{0}^{t} f(z) dz \right]_{0}^{x} \): - At \( t = x \), it is \( x \int_{0}^{x} f(z) dz \).- At \( t = 0 \), the product is zero because \( t = 0 \).Therefore, the boundary term is \( x \int_{0}^{x} f(z) dz \).

Simplify the Remaining Integral

The remaining integral is \( \int_{0}^{x} t f(t) dt \). We rewrite the expression:\[ x \int_{0}^{x} f(z) dz - \int_{0}^{x} t f(t) dt = \int_{0}^{x} f(t) (x-t) \, dt. \]This matches the given expression and concludes the derivation.

Key concepts

Double Integrals

Double integrals extend the concept of an integral to functions of two variables. They are essentially a way to accumulate values over a region in the plane. Consider the function \( f(x, y) \), a double integral \( \iint f(x, y) \, dx \, dy \) helps us find the accumulated result over a specified region.In the exercise we dealt with, we primarily focused on the manipulation of double integrals by breaking them down into iterated integrals. Essentially, an iterated integral means integrating one variable at a time in succession. This process helped meet the goal of converting the original double integral into a more manageable form using integration by parts.When you compute a double integral, you usually do it using two iterated integrals: integrate first with respect to one variable, and then with respect to the other. This method is valuable in situations where the function can be separated into individual variable components.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, providing a way to evaluate definite integrals. It tells us that if you have a continuous function \( f \) that has an antiderivative \( F \) on the interval \([a, b]\), then:

• The derivative of the integral of \( f \) from \( a \) to \( x \) with respect to \( x \) is simply \( f(x) \).
• Thus, \( \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \).
• Moreover, \( \int_{a}^{b} f(t) \, dt = F(b) - F(a) \).

In our exercise solution, this theorem allowed us to differentiate under the integral sign, showing that the inner integral respects this rule as \( f(t) \) emerged upon differentiating the integral from 0 to \( t \). This relationship emphasized why integration by parts was applicable and highlighted how derivatives and integrals are interconnected.

Antiderivatives

Antiderivatives, also known as indefinite integrals, represent the reverse process of differentiation. Finding the antiderivative of a function means determining another function whose derivative is the original function.In our integration problem, we used antiderivatives to simplify the double integral. The function \( f(z) \) is integrated from 0 to \( t \), which essentially means we are finding the antiderivative of \( f \) with respect to \( z \). This setup is necessary before applying integration by parts, as antiderivatives form the basis of the integral used in the method.Understanding antiderivatives is crucial for solving more complex integration problems as they lay the groundwork for integral calculations. An antiderivative makes it possible to evaluate a definite integral, aiding in the transformation of expressions into solvable formats. As seen, they play a pivotal role whenever integration by parts or similar techniques are required to manipulate or simplify integrals for further analysis.