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 Read more

Steps to Solve

1. 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 ]

2. 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} ]

3. 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.

4. 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 ]

5. 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.