1. Solve the following questions: a) Determine the integral via substitution method: ∫ 12x²(x³ + 2)dx b) Find the integral for y = ∫ (x^{1/2} + 3x^{1/2})dx given the initial condi... 1. Solve the following questions: a) Determine the integral via substitution method: ∫ 12x²(x³ + 2)dx b) Find the integral for y = ∫ (x^{1/2} + 3x^{1/2})dx given the initial condition that y = x = 0 2. Solve the following questions by Integration by parts: a) ∫ x²e^{2x}dx b) ∫ 16xe^{(x+9)}dx 3. Evaluate the Taylor Series for f(x) = x³  10x² + 6 at x = 3
Understand the Problem
The question consists of multiple mathematical problems involving integrals and Taylor series. The user needs to solve integrals using substitution and integration by parts, and evaluate a Taylor series for a given function.
Answer
1. \( 2(x^3 + 2)^2 + C \) 2. \( y = \frac{2}{3}x^{3/2} + 6x^{1/2} \) 3. Use \( = \frac{1}{2} x^2 e^{2x}  \int x e^{2x} dx \) 4. \( = 16xe^{(x+9)}  16e^{(x+9)} + C \) 5. \( f(3) = 57 \)
Answer for screen readers

( 2(x^3 + 2)^2 + C )

( y = \frac{2}{3}x^{3/2} + 6x^{1/2} )

For ( \int x^2 e^{2x} dx ): Continue with integration by parts; initial result ( = \frac{1}{2} x^2 e^{2x}  \int x e^{2x} dx )

For ( \int 16x e^{(x+9)} dx ): ( = 16xe^{(x+9)}  16e^{(x+9)} + C )

( f(3) = 57 )
Steps to Solve

Integral via Substitution To solve the integral ( \int 12x^2(x^3 + 2)dx ), we can use the substitution method. Let: [ u = x^3 + 2 ] Then, we find the differential: [ du = 3x^2 dx \quad \Rightarrow \quad dx = \frac{du}{3x^2} ] Now substituting in the integral: [ 12x^2(x^3 + 2)dx = 12x^2 u \cdot \left(\frac{du}{3x^2}\right) = 4u , du ] The integral becomes: [ \int 4u , du = 4 \cdot \frac{u^2}{2} + C = 2u^2 + C ] Substituting back for ( u ): [ 2(x^3 + 2)^2 + C ]

Finding the Integral with Initial Conditions We have the integral ( y = \int (x^{1/2} + 3x^{1/2})dx ). This can be solved by integrating each term: [ y = \int x^{1/2} dx + \int 3x^{1/2} dx ] Evaluating the integrals: [ \int x^{1/2} dx = \frac{2}{3}x^{3/2} + C_1, \quad \int 3x^{1/2} dx = 6x^{1/2} + C_2 ] Combining, we have: [ y = \frac{2}{3}x^{3/2} + 6x^{1/2} + C ] Now, using the initial condition ( y(0) = 0 ): [ 0 = \frac{2}{3}(0)^{3/2} + 6(0)^{1/2} + C \quad \Rightarrow \quad C = 0 ] Hence, [ y = \frac{2}{3}x^{3/2} + 6x^{1/2} ]

Integration by Parts for ( \int x^2 e^{2x} dx ) Using integration by parts where ( u = x^2 ) and ( dv = e^{2x} dx ): [ du = 2x , dx, \quad v = \frac{1}{2} e^{2x} ] Applying integration by parts: [ \int u , dv = uv  \int v , du ] Thus, [ \int x^2 e^{2x} dx = x^2 \cdot \frac{1}{2}e^{2x}  \int \frac{1}{2}e^{2x}(2x) , dx ] Simplifying: [ = \frac{1}{2}x^2 e^{2x}  \int x e^{2x} dx ] The remaining integral can again be solved with integration by parts.

Integration by Parts for ( \int 16x e^{(x+9)} dx ) Here, set ( u = x ) and ( dv = 16 e^{(x+9)} dx ): [ du = dx, \quad v = 16 e^{(x+9)} ] The integral becomes: [ \int 16xe^{(x+9)}dx = 16xe^{(x+9)} + \int 16e^{(x+9)}dx ] Which simplifies to: [ 16xe^{(x+9)}  16e^{(x+9)} + C ]

Evaluating the Taylor Series for ( f(x) = x^3  10x^2 + 6 ) at ( x = 3 ) To evaluate the Taylor series at ( x = 3 ), we find ( f(3) ): [ f(3) = 3^3  10 \cdot 3^2 + 6 = 27  90 + 6 = 57 ] We can also compute higher derivatives if needed, but this gives the function's value at the specified point.

( 2(x^3 + 2)^2 + C )

( y = \frac{2}{3}x^{3/2} + 6x^{1/2} )

For ( \int x^2 e^{2x} dx ): Continue with integration by parts; initial result ( = \frac{1}{2} x^2 e^{2x}  \int x e^{2x} dx )

For ( \int 16x e^{(x+9)} dx ): ( = 16xe^{(x+9)}  16e^{(x+9)} + C )

( f(3) = 57 )
More Information
The problems involve various techniques of integration such as substitution and integration by parts, as well as Taylor series evaluation. Understanding these concepts is crucial in calculus.
Tips
 Forgetting to substitute back for the variable after integrating.
 Not simplifying the boundaries after using integration by parts, especially if they require subsequent integrations.
 Failing to account for constants when integrating.
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