solve this problem---Extraction Content: Question Stem: The image displays the following mathematical expression: Integral from 0 to infinity of sqrt(x) * e^(-x^3) dx Mathematical Formulas: $\int_0^\infty \sqrt{x} e^{-x^3} dx$ Plain text representation of the formula: Integral from 0 to infinity of (sqrt(x) * e^(-x^3)) dx

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We need to evaluate the improper integral from 0 to infinity of square root x times e to the negative x cubed. This integral has several key characteristics: it has an infinite upper limit, contains a square root function and exponential decay term. The rapid decay of e to the negative x cubed ensures convergence, but standard integration techniques won't work directly. We'll need to use substitution methods to transform this into a more manageable form. To solve this integral, we'll use the substitution u equals x cubed. This gives us du equals 3 x squared dx, so dx equals du over 3 x squared. Since x equals u to the one-third power, we have x squared equals u to the two-thirds power. The square root of x becomes u to the one-sixth power. The integration limits remain the same: when x equals zero, u equals zero, and when x approaches infinity, u approaches infinity. After substitution, our integral transforms to the integral from zero to infinity of u to the one-sixth times e to the negative u times du over 3 u to the two-thirds. Now we simplify the transformed integral. We have u to the one-sixth divided by 3 u to the two-thirds times e to the negative u. Combining the exponents: u to the one-sixth minus two-thirds equals u to the negative one-half. So our integral becomes one-third times the integral from zero to infinity of u to the negative one-half times e to the negative u du. We can rewrite this as one-third times the integral of u to the one-half minus one times e to the negative u du. This matches the gamma function form, where gamma of n equals the integral from zero to infinity of t to the n minus one times e to the negative t dt. Our integral equals one-third times gamma of one-half. Now we evaluate gamma of one-half. This is a well-known result: gamma of one-half equals square root of pi. This can be derived from the Gaussian integral, which states that the integral from negative infinity to positive infinity of e to the negative x squared dx equals square root of pi. Applying this to our problem: our integral equals one-third times gamma of one-half, which equals one-third times square root of pi, giving us square root of pi over 3. Therefore, the integral from zero to infinity of square root x times e to the negative x cubed dx equals square root of pi over 3. Let's solve this integral step by step. We need to evaluate the integral from 0 to infinity of square root of x times e to the negative x cubed dx. This is an improper integral with an exponential decay term. We'll use substitution. Let u equal x cubed. Then du equals 3 x squared dx, and x equals u to the one-third power. The differential becomes dx equals du divided by 3 u to the two-thirds power. We can rewrite square root of x as u to the one-sixth power, and e to the negative x cubed becomes e to the negative u. Now let's transform the integral. After substitution, we get u to the one-sixth power times e to the negative u, divided by 3 u to the two-thirds power. Factoring out one-third and combining the u terms, we get u to the power of one-sixth minus two-thirds, which equals negative one-half. This gives us the gamma function of one-half. The gamma function is defined as the integral from 0 to infinity of t to the s minus 1 power times e to the negative t dt. A key result is that gamma of one-half equals square root of pi. Therefore, one-third times gamma of one-half equals square root of pi divided by 3. This gives us our final answer. Let's verify our answer of square root of pi over 3. The behavior makes sense: as x approaches zero, square root of x approaches zero, and as x approaches infinity, e to the negative x cubed approaches zero very rapidly, ensuring the integral converges. The numerical value is approximately 0.626, which matches the area under our curve. Our solution method involved three key steps: first, the substitution u equals x cubed; second, transforming to gamma function form; and third, evaluating using the known result that gamma of one-half equals square root of pi. Therefore, the integral from zero to infinity of square root x times e to the negative x cubed dx equals square root of pi over 3.