Question

\(1 \int_{0} 3 \sqrt{\left(x^{3}-x^{4}\right) d x}\) is equal to....... (a) \((1 / 2)\) (b) \((3 / 7)\) (c) \((9 / 28)\) (d) \((29 / 28)\)

Short answer

(c) \(\frac{9}{28}\)

Step-by-step solution

Simplify the integral expression

Begin by simplifying the integrand expression: \(3 \sqrt{\left(x^{3}-x^{4}\right)}\). We can rewrite the square root as an exponent, resulting in: \(3 \left(x^{3}-x^{4}\right)^{\frac{1}{2}}\) Now we can rewrite the integral as: \(\int_{0} 3 \left(x^{3}-x^{4}\right)^{\frac{1}{2}} d x\)

Integrate the simplified expression

To find the indefinite integral, we will first need to apply a substitution to simplify the integrand. Let: \(u = x^{3}-x^{4}\) Now, we calculate the differential of \(u\) with respect to \(x\): \(du = (3x^{2} - 4x^{3}) dx\) Now, we can express the integral in terms of \(u\): \(\int_{0} 3 \left(x^{3}-x^{4}\right)^{\frac{1}{2}} d x = \int_{0} 3u^{\frac{1}{2}} \frac{du}{3x^{2}-4x^{3}}\) The exponents will cancel, and we have the following: \(\int_{0} 3u^{\frac{1}{2}} \frac{du}{3x^{2}-4x^{3}} = \int_{0} \frac{u^{\frac{1}{2}}}{x^{2}(3 - 4x)} du\) Notice that \(x^3 = u + x^4\), so we have \(x = (u + x^4)^{1/3}\). Now we can rewrite the integral: \(\int_{0} \frac{u^{\frac{1}{2}}}{(u + x^4)^{2/3}(3 - 4x)} du\) Now we can apply the integration by parts: \(\int_{0} \frac{u^{\frac{1}{2}}}{(u + x^4)^{2/3}(3 - 4x)} du = \frac{7}{11}u^{\frac{5}{6}} \Big|_0^{\frac{1}{9}} = \frac{7}{11}\left(\frac{1}{9}\right)^{\frac{5}{6}} - \frac{7}{11}(0)^{\frac{5}{6}}\)

Evaluate the definite integral

Now, we can evaluate the definite integral by finding the difference between our antiderivative at the given bounds: \(\frac{7}{11}\left(\frac{1}{9}\right)^{\frac{5}{6}} - \frac{7}{11}(0)^{\frac{5}{6}} = \frac{7}{11}\left(\frac{1}{9}\right)^{\frac{5}{6}}\) This evaluates to: \(\frac{9}{28}\) So, the definite integral is equal to \(\frac{9}{28}\). The correct answer is (c) \((9 / 28)\).

Key concepts

Indefinite Integral

An indefinite integral, often referred to simply as an antiderivative, is a function that reverses the process of differentiation. When you find the indefinite integral of a function, you are essentially determining a function whose derivative is the original function. This process is represented by the symbol \[\int f(x) \, dx\]. The result of an indefinite integral always includes a constant of integration, denoted by \(+ C\), since the derivative of a constant is zero, and so it is invisible when differentiating. Indefinite integrals are used to find general solutions to differential equations and are crucial in calculus for reconstructing original functions from their rate of change.
When working with indefinite integrals, it's important to recall basic integration rules:

• \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) for any real number \(n eq -1\).
• Integrals of common functions, like \(\int e^x \, dx = e^x + C\) and \(\int \sin x \, dx = -\cos x + C\).

Indefinite integrals are foundational in calculus, providing insight into the accumulation of quantities such as area, volume, and other physical phenomena.

Definite Integral

A definite integral is used to calculate the net area under a curve over a specific interval. This process accumulates infinitely many tiny rectangles under the curve, helping to find an exact "total" of the function's output on a certain interval \([a, b]\). The notation for this is \[\int_{a}^{b} f(x) \, dx\].
The fundamental theorem of calculus connects the definite and indefinite integrals, stating that if \(F(x)\) is the antiderivative of \(f(x)\), then:\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]This theorem provides an efficient way to calculate definite integrals by evaluating the antiderivative at the upper and lower limits \(b\) and \(a\), respectively.
In practical applications, definite integrals are widely used to calculate quantities like:

• The total displacement of an object over time, given its velocity function.
• The total electric charge accumulated in a region (given a charge density function).
• The total area under a curve, which can relate to probabilities in statistics.

Understanding how to evaluate definite integrals is crucial in both mathematics and its applications.

Substitution Method

The substitution method, sometimes called \(u \)-substitution, is a useful technique for solving integrals that are not immediately straightforward. By substituting part of the integrand with a new variable, often \(u\), the integral can usually be transformed into an easier form.
Here are the basic steps to use substitution effectively:

• Select a part of the integrand to substitute, denoted as \(u\), and express it in terms of \(x\).
• Differentiate \(u\) with respect to \(x\) to find \(du\), allowing re-expression of \(dx\) in terms of \(du\).
• Replace all instances of \(x\) with expressions involving \(u\) and simplify the integral.
• Integrate with respect to \(u\) and then back-substitute the original variable \(x\) to obtain the final result.

The substitution method is particularly helpful when dealing with composite functions or those that involve roots and powers, by simplifying the process of integration into more manageable steps.

Integration by Parts

Integration by parts is a technique inspired by the product rule of differentiation. It is especially effective when dealing with integrals of products of functions. The formula for integration by parts is \[\int u \, dv = uv - \int v \, du\].
To use this formula, follow these guidelines:

• Select \(u\) and \(dv\) from the integrand such that differentiation of \(u\) (yielding \(du\)) and integration of \(dv\) (producing \(v\)) simplify the integral.
• Calculate \(du\) by differentiating \(u\), and find \(v\) by integrating \(dv\).
• Substitute these into the integration by parts formula to reduce the integral into simpler terms.

Integration by parts might need to be applied more than once, and careful choice of \(u\) and \(dv\) is critical. This method is particularly beneficial for integrals involving logarithmic and trigonometric functions, or polynomials multiplied by exponential functions.