Question

consider the definite integral $int_{0}^{1} xe^{4x^{2}+b}dx$, where $b$ is a real number. your answers below may be in terms of $b$.
a) this can be transformed using the substitution
$u = $

b) this gives $du = $

c) performing the substitution in terms of $u$ and changing the limits of integration gives the integral
$int_{square}^{square}$

d) evaluate the integral and simplify.

Explanation:

Step1: Choose substitution for exponent

Let $u = 4x^2 + b$

Step2: Compute derivative of u

$du = \frac{d}{dx}(4x^2 + b)dx = 8x dx$

Step3: Rewrite integral for substitution

Rearrange $du=8x dx$ to get $x dx = \frac{1}{8}du$.
Find new limits:
When $x=0$, $u=4(0)^2 + b = b$
When $x=1$, $u=4(1)^2 + b = 4 + b$
Substitute into original integral:
$\int_{b}^{4+b} e^u \cdot \frac{1}{8}du$

Step4: Evaluate the definite integral

Integrate $\frac{1}{8}e^u$:
$\frac{1}{8}e^u \bigg|_{b}^{4+b} = \frac{1}{8}(e^{4+b} - e^b)$

Answer:

a) $4x^2 + b$
b) $8x dx$
c) Lower limit: $b$, Integrand: $\frac{1}{8}e^u$, Upper limit: $4+b$
d) $\frac{1}{8}e^b(e^4 - 1)$ (or $\frac{e^{4+b} - e^b}{8}$)