# Use integration by parts to find the given integral. Use symbolic notation and fractions where needed. Integra sign has 1 on top and 0 on bottom....2x(e^(-3x)  + e^(-x)) dx

You need to evaluate the definite integral, such that:

int_0^1 2x(e^(-3x) + e^(-x))dx

You need to open the brackets such that:

int_0^1 (2xe^(-3x) + 2xe^(-x))dx

You need to use the property of linearity of integral, such that:

int_0^1 (2xe^(-3x) + 2xe^(-x))dx = int_0^1 (2xe^(-3x))dx + int_0^1 (2xe^(-x))dx

You need to solve the integrals, using integration by parts, such that:

int_a^b udv = uv|_a^b - int_a^b vdu

Considering u = x and dv = e^(-3x)dx yields:

u = x => du = dx

dv = e^(-3x)dx => v = -(e^(-3x))/3

int_0^1 (2xe^(-3x))dx = 2int_0^1 (xe^(-3x))dx

int_0^1 (xe^(-3x))dx = -x(e^(-3x))/3|_0^1 + (1/3)int_0^1 e^(-3x)dx

int_0^1 (xe^(-3x))dx = -x(e^(-3x))/3|_0^1 - (1/9)e^(-3x)|_0^1

2int_0^1 (xe^(-3x))dx = 2(-x(e^(-3x))/3|_0^1 - (1/9)e^(-3x)|_0^1)

2int_0^1 (xe^(-3x))dx = 2(-1*e^(-3)/3 - e^(-3)/9 + e^0/9)

2int_0^1 (xe^(-3x))dx = -2/(3e^3) - 1/(9e^3) + 1/9

2int_0^1 (xe^(-3x))dx = -7/(9e^3) + 1/9

2int_0^1 (xe^(-3x))dx = (e^3 - 7)/(9e^3)

Solving the integral int_0^1 (2xe^(-x))dx using parts yields:

u = x => du = dx

dv = e^(-x)dx => v = -e^(-x)

int_0^1 (2xe^(-x))dx = 2int_0^1 (xe^(-x))dx

2int_0^1 (xe^(-x))dx = 2(-xe^(-x)|_0^1 - e^(-x)|_0^1)

2int_0^1 (xe^(-x))dx = 2(-1/e - 1/e + e^0) 2int_0^1 (xe^(-x))dx = -4/e + 1 => 2int_0^1 (xe^(-x))dx = (e - 4)/e

int_0^1 (2xe^(-3x) + 2xe^(-x))dx = (e^3 - 7)/(9e^3) +(e - 4)/e

Hence, evaluating the given definite integral using parts, yields int_0^1 (2xe^(-3x) + 2xe^(-x))dx = (e^3 - 7)/(9e^3) +(e - 4)/e.

## See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Approved by eNotes Editorial