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Chapter 7: Problem 20

Evaluate the following integrals. $$\int_{0}^{2} \frac{x(3 x+2)}{\sqrt{x^{3}+x^{2}+4}} d x$$

Short Answer

Expert verified
**Question:** Find the value of the following integral: $$\int_{0}^{2} \frac{x(3 x+2)}{\sqrt{x^{3}+x^{2}+4}} d x$$ **Answer:** The value of the given integral is 16.

Step by step solution

01

Simplify the integrand

First, let's simplify the given function: $$\frac{x(3 x+2)}{\sqrt{x^{3}+x^{2}+4}}$$ There is not much to simplify in the given function, so we leave it as it is.
02

Choose a substitution

To make the integration process easier, let's choose a substitution for the expression inside the square root. Let: $$ u = x^3 + x^2 + 4 $$ Then, differentiate u with respect to x: $$ \frac{d u}{d x} = 3x^2 + 2x $$ Now, solve for \(dx\): $$ d x = \frac{d u}{3x^2 + 2x} $$ Our integral now becomes: $$ \int_{0}^{2} \frac{x(3 x+2)}{\sqrt{u}} \cdot \frac{d u}{3x^2 + 2x}$$
03

Rewrite the integral in terms of u

Substitute the expressions we found in Step 2 into the integral: $$ \int_{0}^{2} \frac{x(3 x+2)}{\sqrt{u}} \cdot \frac{d u}{3x^2 + 2x}$$ Notice that the \((3x^2 + 2x)\) cancels with the term in the denominator of \(dx\), resulting in: $$ \int_{0}^{2} \frac{x}{\sqrt{u}} d u $$ Remember that we also need to change the limits of integration. Since we have \(u = x^3 + x^2 + 4\), the new limits of integration are: For \(x=0\): $$ u = 0^3 + 0^2 + 4 = 4 $$ For \(x=2\): $$ u = 2^3 + 2^2 + 4 = 8 + 4 + 4 = 16 $$ Our integral now becomes: $$ \int_{4}^{16} \frac{x}{\sqrt{u}} d u $$
04

Integrate with respect to u

To find the antiderivative, we can use power rule: $$\int \frac{x}{\sqrt{u}} d u = 2x \sqrt{u} + C $$ Now, we can substitute back x in terms of u to get the result in terms of x: $$ 2x \sqrt{x^3 + x^2 +4}+ C $$
05

Apply limits of integration

Using the Fundamental Theorem of Calculus, we can find the value of the definite integral by evaluating the antiderivative at the bounds and subtracting: $$ \int_{4}^{16} \frac{x}{\sqrt{u}} d u = \left[ 2x \sqrt{x^3 + x^2 +4} \right]_{4}^{16} $$ Evaluate at the upper bound: $$ 2(2) \sqrt{2^3 + 2^2 + 4} = 4 \sqrt{16} = 16 $$ Evaluate at the lower bound: $$ 2(0) \sqrt{0^3 + 0^2 + 4} = 0 $$ Subtract the two values: $$ 16 - 0 = 16 $$ The value of the given integral is: $$\int_{0}^{2} \frac{x(3 x+2)}{\sqrt{x^{3}+x^{2}+4}} d x = 16 $$

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