Question

Lösen Sie durch Substitution a) \(\int_{0}^{1}(5 x-4)^{3} d x\) b) \(\int_{1}^{2} \frac{d x}{\sqrt{7-3 x}}\) Substituieren Sie: \(u=7-3 x\). c) \(\int_{1}^{\frac{3}{2}} \sin \left(\pi x+\frac{5 \pi}{2}\right) d x\) d) \(\int_{-1}^{1} x^{2} \sqrt{2 x^{3}+4} d x\)

Short answer

a) \( \frac{257}{20} \) b) \( \frac{2}{3} \) c) \( - \frac{1}{\pi} \) d) \( \frac{2 \sqrt{6} - 2 \sqrt{2}}{3} \)

Step-by-step solution

Teil a – Bestimmen Sie die Substitution

Für das Integral \ \( \int_{0}^{1}(5 x-4)^{3} \, d x \), setzen Sie \(u = 5x - 4\). Berechnen Sie dann \( \frac{d u}{d x} \).

Teil a – Berechnung des Integrals

Da \( \frac{d u}{d x} = 5 \), ist \( \,dx = \frac{du}{5} \). Ersetzen Sie die Integrationsgrenzen: Wenn \(x = 0\), ist \(u = -4\); und wenn \(x = 1\), ist \(u = 1\). Das Integral wird dann zu \( \int_{-4}^{1} \frac{u^{3}}{5} \, d u \). Dies vereinfacht sich zu \( \frac{1}{5} \int_{-4}^{1} u^{3} \, d u \).

Teil a – Abschluss der Berechnung

Berechnen Sie das Integral von \(u^{3}\): \( \int u^{3} \, d u = \frac{u^{4}}{4} \). Setzen Sie die Grenzen ein: \[ \frac{1}{5} \left[ \frac{u^{4}}{4} \right]_{-4}^{1} = \frac{1}{20} \left(1 - (-256)\right) = \frac{1}{20} \left(257\right) = \frac{257}{20} \]. Der Wert des Integrals ist \( \frac{257}{20} \).

Teil b – Substitution wählen

Für das Integral \( \int_{1}^{2} \frac{d x}{\sqrt{7-3x}} \), setzen Sie \(u = 7 - 3x\). Berechnen Sie \( \frac{d u}{d x} \).

Teil b – Berechnung des Integrals

Da \( \frac{d u}{d x} = -3 \), ist \( d x = \frac{d u}{-3} \). Ersetzen Sie die Integrationsgrenzen: Wenn \(x = 1\), ist \(u = 4\); und wenn \(x = 2\), ist \(u = 1\). Das Integral wird dann zu \( -\frac{1}{3} \int_{4}^{1} u^{-\frac{1}{2}} \, d u \).

Teil b – Abschluss der Berechnung

Berechnen Sie das Integral von \(u^{-1/2}\): \( \int u^{-1/2} \, d u = 2u^{1/2} \). Setzen Sie die Grenzen ein: \[ -\frac{1}{3} \left[2u^{1/2}\right]_{4}^{1} = -\frac{2}{3} \left(1 - 2\right) = -\frac{2}{3} \left(-1\right) = \frac{2}{3} \]. Der Wert des Integrals ist \(\frac{2}{3}\).

Teil c – Bestimmen Sie die Substitution

Für das Integral \( \int_{1}^{\frac{3}{2}} \sin \left(\pi x + \frac{5 \pi}{2}\right) \, d x \), setzen Sie \(u = \pi x + \frac{5 \pi}{2}\). Berechnen Sie \( \frac{d u}{d x} \).

Teil c – Berechnung des Integrals

Da \( \frac{d u}{d x} = \pi \), ist \( \, dx = \frac{d u}{\pi} \). Ersetzen Sie die Integrationsgrenzen: Wenn \( x = 1 \), ist \( u = \frac{7 \pi}{2} \); und wenn \( x = \frac{3}{2} \), ist \( u = 4 \pi \). Das Integral wird dann zu: \( \frac{1}{\pi} \int_{\frac{7 \pi}{2}}^{4 \pi} \sin(u) \, d u \).

Teil c – Abschluss der Berechnung

Berechnen Sie das Integral von \( \sin(u) \): \( \int \sin(u) \, d u = - \cos(u) \). Setzen Sie die Grenzen ein: \[ \frac{1}{\pi} \left[ - \cos(u) \right]_{\frac{7 \pi}{2}}^{4 \pi} = - \frac{1}{\pi} \left( \cos(4 \pi) - \cos \left( \frac{7 \pi}{2} \right)\right) = - \frac{1}{\pi} \left( 1 - 0 \right) = - \frac{1}{\pi} \]. Der Wert des Integrals ist \( - \frac{1}{\pi} \).

Teil d – Bestimmen Sie die Substitution

Für das Integral \( \int_{-1}^{1} x^{2} \sqrt{2x^{3} + 4} \, d x \), setzen Sie \(u = 2x^{3} + 4\). Berechnen Sie \( \frac{d u}{d x} \).

Teil d – Berechnung des Integrals

Da \( \frac{d u}{d x} = 6x^{2} \), ist \( d x = \frac{d u}{6 x^{2}} \). Ersetzen Sie die Integrationsgrenzen: Wenn \( x = -1 \), ist \( u = 2 \); und wenn \( x = 1 \), ist \( u = 6 \). Das Integral wird dann zu: \( \frac{1}{6} \int_{2}^{6} \sqrt{u} \, d u \).

Teil d – Abschluss der Berechnung

Berechnen Sie das Integral von \( \sqrt{u} \): \( \int \sqrt{u} \, d u = \frac{2}{3} u^{3 /2} \). Setzen Sie die Grenzen ein: \[ \frac{1}{6} \left[ \frac{2}{3} u^{3/2} \right]_{2}^{6} = \frac{1}{9} \left( 6^{3/2} - 2^{3/2} \right) = \frac{1}{9} \left( 6 \sqrt{6} - 2 \sqrt{2} \right) \]. Der Wert des Integrals ist \( \frac{2 \sqrt{6} - 2 \sqrt{2}}{3} \).

Key concepts

Substitution

The method of substitution is a powerful tool in integral calculus for simplifying functions before integrating. Substitution helps in transforming complicated integrals into simpler ones, making them easier to manage.
To use substitution, we follow these steps:

• Choose a substitution variable, say, u. This should be a function of x that simplifies the given integral. For example, if we have \(\int_{0}^{1} (5x-4)^{3} \ \ dx\), we can set \(\ u = 5x - 4\).
• Calculate the differential du. In our example, \(\du = 5 dx\)
• Figuring out the new limits after substitution. For instance: when \(\x = 0\), \(\u = -4\)

By transforming the integral into a simpler function of u, we can proceed with integration easily.

Definites Integral

A definite integral is used to calculate the area under a curve between two specified points. It evaluates the integral with limits, thus giving a specific numerical result, unlike an indefinite integral which includes a constant of integration.
To solve a definite integral, follow these steps:

• Identify the limits of integration; for instance, in \(\int_{1}^{2} \/ \sqrt{7-3x} \ dx\), the limits are from 1 to 2.
• Apply substitution if needed to simplify the integral. If we let \(\u = 7 - 3x\), we need to change the limits accordingly.
• Integrate the simpler function within the new limits, then substitute back to obtain the answer in terms of the original variable if necessary.

For example, after substitution for \(\int_{1}^{2} \/ \sqrt{7-3x} \ dx\), the new integral becomes \(\-\ \int_{4}^{1} u^{-\frac{1}{2}} du\). Finally, calculate the definite integral within these new limits.

Integrationstechnik

Integration techniques are methods used to solve integrals that cannot easily be evaluated through basic integration rules. One such technique is integration by substitution.
Some essential integration techniques include:

• Substitution: Used when the integral includes a composite function, simplifying it into a form that is easier to integrate.
• Integration by parts: Often used when the product of two functions is to be integrated.
• Partial fractions: Particularly useful for rational functions.

For the exercises given, substitution is a commonly used method, transforming a more complex integral into an easier one. For example, in solving
\(\int_{1}^{\frac{3}{2}} \/ \sin (\pi x + \frac{5 \pi}{2}) \ dx\), we simplify by letting \(\u = \pi x + \frac{5 \pi}{2}\).
Recognize when to apply each technique will make solving integrals more straightforward.

Mathematische Analyse

Mathematical analysis focuses on understanding changes, accumulation, and the relationships between different measurable quantities. Integral calculus, a branch of mathematical analysis, centers on integrals and their applications.
Important concepts in mathematical analysis with respect to integration include:

• The Fundamental Theorem of Calculus, which connects differentiation and integration.
• Definite and indefinite integrals and their uses for calculating areas and antiderivatives.
• Techniques for evaluating complex integrals, such as substitution and integration by parts.

Each concept builds a foundation for more advanced mathematical studies. For instance, the solution to the integral
\ <\int_{-1}^{1} x^{2} \/ \sqrt{2x^{3} + 4} \ dx\ involves using the substitution \(\u = 2x^{3} + 4, \) streamlining the problem into a simpler integral.
Developing proficiency in these concepts ensures a robust grasp of mathematical analysis.