Question

We want to compute the integral ∫ (2ˣ²+3x+4)ᵉ²ˣdx using two (and only two) integration by parts. This means that you have to make a good choice of f and g for your two integration by parts. For the first intagration by parts, you should choose [f(x),g

(x)]= Warning: for this question, you must use strict calculator notations and also include the square brackets in your answer; in particular, you must use

Answer 1

To solve ∫ (2x²+3x+4)e²x dx, we use integration by parts twice. For the first round, choose f(x) as (2x²+3x+4) and g'(x) as e²x. Differentiate f(x) and integrate g'(x), then apply integration by parts a second time.

Step-by-step explanation:

To compute the integral ∫ (2x²+3x+4)e²xdx using two applications of integration by parts, we must choose the functions to differentiate and integrate wisely. The integrand is a function multiplied by an exponential function, which is a common scenario for using integration by parts. When choosing our functions, we typically let u be the function that simplifies when differentiated, and dv be the function that remains manageable when integrated.

For the first application of integration by parts, we can choose:

• f(x) as the polynomial part (2x²+3x+4)
• g'(x) as the exponential part e²x

This choice allows us to carry out the integration by parts formula:

∫ u dv = uv - ∫ v du

After choosing our f(x) and g'(x), we would then differentiate f(x) and integrate g'(x) to find df/dx and g(x), respectively. A second integration by parts would then be done with our new functions, following the same reasoning to simplify the integral further.