We are asked to evaluate the definite integral of the function $y = x^2 + 4x + 3$. The integral is given as: $\int_{-3}^{0} (x^2 + 4x + 3) \, dx$ The "x axis" likely refers to the lower limit of the integration being the x-intercept of the curve to the left of the y-axis.

1. Problem Description

We are asked to evaluate the definite integral of the function $y = x^2 + 4x + 3$. The integral is given as:

$\int_{-3}^{0} (x^2 + 4x + 3) \, dx$

The "x axis" likely refers to the lower limit of the integration being the x-intercept of the curve to the left of the y-axis.

2. Solution Steps

First, find the antiderivative of $x^2 + 4x + 3$.

$\int (x^2 + 4x + 3) \, dx = \frac{x^3}{3} + 2x^2 + 3x + C$.

Now, evaluate the definite integral between the limits $-3$ and $0$.

The antiderivative is:

$F(x) = \frac{x^3}{3} + 2x^2 + 3x$

We need to evaluate $F(0) - F(-3)$.

$F(0) = \frac{(0)^3}{3} + 2(0)^2 + 3(0) = 0$

$F(-3) = \frac{(-3)^3}{3} + 2(-3)^2 + 3(-3) = \frac{-27}{3} + 2(9) - 9 = -9 + 18 - 9 = 0$

So,

$\int_{-3}^{0} (x^2 + 4x + 3) \, dx = F(0) - F(-3) = 0 - 0 = 0$.

Note: $x^2 + 4x + 3 = (x+1)(x+3)$.

Therefore $x = -1, -3$ are the roots. So we want the area between -3 and -1 which is negative, and then the area between -1 and 0, which is positive. The area is given by the definite integral.

3. Final Answer

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