Math 266 Calculus For Engineers Project Three: Integration

Question

Math 266 Calculus For Engineers Iiproject Threeintegration Methodsthe Math 266: Calculus for Engineers II Project Three focuses on applying various integration techniques, such as substitution methods, utilization of integration tables, and computer algebra systems. The key task involves evaluating a complex integral involving polynomials and applying appropriate integration strategies to obtain an exact symbolic result. Your assignment is to analyze the given integral: âˆ« (1.39t4 + 0.84t3 + 0.98t2 + 1.13t + 0.55t5 + 1.57t3 + 0.6t) dt First, determine how many fractions (or terms) are present after expanding and simplifying the integrand. For each term, decide whether it can be integrated directly, or if it requires substitution—specify whether it’s a u-substitution or a trigonometric substitution—and perform the necessary steps. If any term is best integrated using an integral table, identify the general form and final result. Once all terms are integrated, assemble the full indefinite integral solution. Then, use a symbolic calculation tool (such as WolframAlpha, Maple, or Mathematica) to verify your results. Compare the computer-generated symbolic result with your manual solution, noting whether they are equal or merely equivalent (i.e., differ by a constant). Show all necessary work, including the justification for choosing each method, substitution steps, and the final expressions. ---

Dr. Jack HW Helper · Accepted Answer

The integral provided for this calculus project is: $$ \int (1.39t^4 + 0.84t^3 + 0.98t^2 + 1.13t + 0.55t^5 + 1.57t^3 + 0.6t) dt $$ The first step involves simplifying the integrand by combining like terms. We notice that $ 0.84t^3 $ and $ 1.57t^3 $ can be summed, which results in a unified expression: $$ 1.39t^4 + (0.84 + 1.57)t^3 + 0.98t^2 + (1.13 + 0.6)t + 0.55t^5 $$ $$ = 1.39t^4 + 2.41t^3 + 0.98t^2 + 1.73t + 0.55t^5 $$ Thus, the simplified polynomial integrand becomes: $$ 0.55t^5 + 1.39t^4 + 2.41t^3 + 0.98t^2 + 1.73t $$ The integral is now: $$ \int \left(0.55t^5 + 1.39t^4 + 2.41t^3 + 0.98t^2 + 1.73t\right) dt $$ This integral is straightforward, as it involves direct power rule integration. Each term can be integrated separately: $$ \int 0.55t^5 dt = 0.55 \cdot \frac{t^6}{6} = \frac{0.55}{6} t^6 $$ $$ \int 1.39t^4 dt = 1.39 \cdot \frac{t^5}{5} = \frac{1.39}{5} t^5 $$ $$ \int 2.41t^3 dt = 2.41 \cdot \frac{t^4}{4} = \frac{2.41}{4} t^4 $$ $$ \int 0.98t^2 dt = 0.98 \cdot \frac{t^3}{3} = \frac{0.98}{3} t^3 $$ $$ \int 1.73t dt = 1.73 \cdot \frac{t^2}{2} = \frac{1.73}{2} t^2 $$ Combining these, the indefinite integral becomes: $$ \frac{0.55}{6} t^6 + \frac{1.39}{5} t^5 + \frac{2.41}{4} t^4 + \frac{0.98}{3} t^3 + \frac{1.73}{2} t^2 + C $$ Numerically, $$ = 0.0917 t^6 + 0.278 t^5 + 0.6025 t^4 + 0.3267 t^3 + 0.865 t^2 + C $$ To verify this solution, I used WolframAlpha's symbolic integration tool, which produced the same form, confirming the accuracy of the manual work. Comparing both methods, the results are equivalent because they differ only by an arbitrary constant of integration, which is standard in indefinite integrals. The precise algebraic form and coefficients match, providing confidence in the solution's correctness. In conclusion, all terms were directly integrable via the power rule because they involved polynomial expressions. No substitution was necessary for this integral. However, in more complex integrals involving roots or trigonometric functions, su

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Math 266 Calculus For Engineers Iiproject Threeintegration Methodsthe

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Math 266 Calculus For Engineers Iiproject Threeintegration Methodsthe

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