$$\int_0^1\left(x^4-9\right)e^xdx=$$

To evaluate the definite integral using a quick, intuitive approach, we can use the standard reduction formula for integrals of the form $$\int P(n, x) e^x dx$$, where $$P(n, x)$$ is a polynomial.

For any polynomial $$P(x)$$, the integral $$\int P(x) e^x dx$$ can be evaluated efficiently using the differential operator shortcut:

$$\int P(x)e^x dx = \left( P(x) - P'(x) + P''(x) - P'''(x) + \dots \right) e^x$$

Let our polynomial be $$P(x) = x^4 - 9$$.

Let us find its successive derivatives:

• $$P(x) = x^4 - 9$$
  
• $$P'(x) = 4x^3$$
  
• $$P''(x) = 12x^2$$
  
• $$P'''(x) = 24x$$
  
• $$P^{(4)}(x) = 24$$
  
• Higher derivatives = $$0$$
  

Applying the shortcut formula, the antiderivative $$F(x)$$ is:

$$F(x) = \left[ (x^4 - 9) - 4x^3 + 12x^2 - 24x + 24 \right] e^x$$

$$F(x) = (x^4 - 4x^3 + 12x^2 - 24x + 15) e^x$$

Now, apply the limits from $$0$$ to $$1$$:

$$\int_0^1 (x^4 - 9) e^x dx = F(1) - F(0)$$

Substitute $$x = 1$$:

$$F(1) = (1 - 4 + 12 - 24 + 15) e^1 = (28 - 28) e = 0$$

Substitute $$x = 0$$:

$$F(0) = (0 - 0 + 0 - 0 + 15) e^0 = 15 \times 1 = 15$$

Subtracting the two values:

$$\int_0^1 (x^4 - 9) e^x dx = 0 - 15 = -15$$

The correct option is B.