Question

Evaluate the following integrals: $$ \int \sqrt{4-3 x-2 x^{2}} d x $$

Short answer

Question: Evaluate the indefinite integral: \(\int \sqrt{4-3x-2x^2} dx\) Answer: \(-\sqrt{2}\left( \frac{1}{2}\left[(x+\frac{3}{4})\sqrt{\left(\frac{3}{4}\right)^2 - (x+\frac{3}{4})^2} + \left(\frac{3}{4}\right)^2 \arcsin{\frac{2(x + \frac{3}{4})}{3}}\right] + C \right)\)

Step-by-step solution

Complete the square for the quadratic expression

We have the expression inside the square root as: $$ 4-3x-2x^2 $$ To complete the square for this quadratic expression, we perform the following steps: - Factor out -2 from the x terms: $$ 4-3x-2x^2 = -2(2+\frac{3}{2}x+x^2) $$ - Add and subtract the square of half of the x term's coefficient (notice we are only changing the expression inside the parenthesis): $$ 2+\frac{3}{2}x+x^2 = \left(x+\frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2 $$ So now we have: $$ -2\left(\left(x+\frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2\right) $$

Perform the substitution

Now, we notice that the expression inside the square root can be simplified using the substitution: $$ u = x + \frac{3}{4} $$ Thus, we have: $$ du = dx $$ Now we change our integral in terms of \(u\): $$ \int \sqrt{-2\left(u^2 - \left(\frac{3}{4}\right)^2\right)} du $$

Simplify and evaluate the integral

Now, the function inside the integral becomes easier to integrate: $$ \int \sqrt{-2\left(u^2 - \left(\frac{3}{4}\right)^2\right)} du = -\sqrt{2} \int \sqrt{u^2 - \left(\frac{3}{4}\right)^2} du $$ This integral is a standard integral of the form \(\int \sqrt{a^2 - x^2} dx\), which has the solution: $$ \frac{1}{2} \left[x \sqrt{a^2 - x^2} + a^2 \arcsin{\frac{x}{a}}\right] + C $$ Using this formula, we can now evaluate our integral in terms of \(u\): $$ -\sqrt{2}\left( \frac{1}{2}\left[u\sqrt{\left(\frac{3}{4}\right)^2 - u^2} + \left(\frac{3}{4}\right)^2 \arcsin{\frac{2u}{3}}\right] + C \right) $$

Substitute back for x

Now that we have the integral in terms of \(u\), we need to substitute back for \(x\) using our original substitution \(u = x + \frac{3}{4}\). The final result is then: $$ -\sqrt{2}\left( \frac{1}{2}\left[(x+\frac{3}{4})\sqrt{\left(\frac{3}{4}\right)^2 - (x+\frac{3}{4})^2} + \left(\frac{3}{4}\right)^2 \arcsin{\frac{2(x + \frac{3}{4})}{3}}\right] + C \right) $$ This is the final expression for the evaluated integral.

Key concepts

Completing the Square

Completing the square is a method used in algebra to transform a quadratic expression into a perfect square trinomial plus or minus a constant. It's central to solving quadratic equations and is an invaluable technique in integral calculus, especially when dealing with integrands involving quadratic expressions.

Here's how it's done in a nutshell:

• Identify the quadratic expression and rearrange it if necessary, so that the terms are in descending power of 'x'.
• Factor out any common coefficients from the 'x' terms.
• Divide the coefficient of the linear 'x' term by 2, square the result, and add and subtract this square within the expression.
• Rewrite the expression to highlight the perfect square trinomial and the constant.

For instance, in the exercise given, we start with the quadratic expression \(4 - 3x - 2x^2\). After factoring out a '-2', adding and subtracting \(\frac{3}{4}\) squared within the parentheses, we obtain a perfect square \(\big(x + \frac{3}{4}\big)^2\) minus \(\big(\frac{3}{4}\big)^2\), which simplifies the integral evaluation process significantly.

When applied to integrals, completing the square enables the use of substitution and often makes the integral standard or simpler to evaluate. This is a key reason why it's an essential concept in integral calculus.

Integral Substitution

Integral substitution, often referred to as u-substitution, is a method for simplifying integrals. It's akin to an algebraic change of variable that transforms a complicated integral into a more manageable form.

Here’s the gist of the process:

• Identify a portion of the integrand (the function being integrated) that can be substituted with a new variable, 'u', to simplify the expression.
• Calculate the differential 'du', and replace the corresponding 'dx' in the integral.
• Rewrite the entire integral in terms of 'u', simplifying where possible.
• After integrating with respect to 'u', substitute the original variables back into the expression to get the result in terms of the original variable.

In our problem, the substitution \(u = x + \frac{3}{4}\) simplified the integrand by creating a recognizable square root form. The differential du conveniently matched dx, indicating a straightforward substitution without additional manipulations. Substitution is crucial for integrals involving algebraic manipulation like completing the square, because it often converts the integrand into a format for which standard integral formulas apply.

Definite Integrals

Definite integrals calculate the net area under a curve bounded by specific limits. These integrals are at the heart of many applications in fields ranging from physics to economics.

Definite integrals follow a few fundamental principles:

• They are evaluated over a closed interval [a, b], with 'a' and 'b' being the lower and upper limits, respectively.
• The fundamental theorem of calculus holds the key to evaluating these integrals—it connects differentiation with integration.
• After finding the indefinite integral (also known as the antiderivative), the definite integral is found by calculating the difference between the values of the antiderivative at the upper and lower limits.

While the exercise provided does not involve a definite integral, understanding how to evaluate indefinite integrals is a precursor to solving definite integrals. Once you have integrated a function indefinitely, as in our exercise, you can apply the limits of integration if needed, to find the specific area under the curve between two points on the x-axis.

Definite integrals express accumulation of quantities and therefore are a fundamental tool in analyzing physical problems related to distance, area, volume, and other quantities that accumulate over an interval.