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
$u = 4x^{2}+b$
b) this gives $du = 8xdx$
c) performing the substitution in terms of $u$ and changing the limits of integration gives the integral
d) evaluate the integral and simplify.
$\frac{1}{8}e^{b}(e^{4}-1)$

Explanation:

Step1: Adjust du for xdx

From $du=8xdx$, solve for $xdx$:
$\frac{1}{8}du = xdx$

Step2: Find new u-limits

At $x=0$: $u=4(0)^2+b = b$
At $x=1$: $u=4(1)^2+b = 4+b$

Step3: Substitute into integral

Replace $xdx$, $4x^2+b$, and limits:
$$\int_{b}^{4+b} \frac{1}{8}e^u du$$

Answer:

The substituted definite integral is $\boldsymbol{\int_{b}^{4+b} \frac{1}{8}e^u du}$