Evaluate the definite integral. \int_{-2}^{3} (-x^{2} - 3x + 7) dx


Evaluate the definite integral.

{eq}\displaystyle\; \int_{-2}^{3} \left(-x^{2} - 3x + 7\right)\,dx {/eq}

Definite Integrals

To solve a definite integral, we use the Fundamental Theorem of Calculus part II,

{eq}\displaystyle \int_a^b f(x)\ dx=F(x)\bigg\vert_a^b=F(b)-F(a),\\ {/eq} where {eq}\displaystyle F(x) \text{ is an antiderivative of } f(x): \frac{d}{dx}(F(x))=f(x). {/eq}

The general antiderivative function is {eq}\displaystyle F(x)+c \text{ because } \frac{d}{dx}(F(x)+c)=f(x), c-\text{ constant}. {/eq}

Integrating, we ma need the following linearity property and formula

{eq}\displaystyle \int_a^b\left[\alpha f(x)+\beta g(x)\right]\ dx= \alpha \int_a^bf(x)\ dx+ \beta \int_a^b g(x)\ dx, a, b, \alpha, \beta -\text{ constants}\\ \displaystyle \int x^n dx=\frac{1}{n+1}x^{n+1}+C, n\neq -1, C - \text{ constant }\\ {/eq}

Answer and Explanation:

To evaluate the definite integral {eq}\displaystyle \displaystyle\; \int_{-2}^{3} \left(-x^{2} - 3x + 7\right)\,dx {/eq}

we will use the linearity properties of the definite integral and the Fundamental Theorem of Calculus Part II, as below.

{eq}\displaystyle \begin{align} \displaystyle\; \int_{-2}^{3} \left(-x^{2} - 3x + 7\right)\,dx&= = -\int_{-2}^{3}x^{2}\ dx - 3 \int_{-2}^{3} x\ dx + 7\int_{-2}^{3}\,dx, &\left[\text{using the linearity of the definite integrals}\right]\\ &= -\frac{x^{3}}{3}\bigg\vert_{-2}^3 - 3 \cdot \frac{x^2}{2}\bigg\vert_{-2}^3 + 7 x\bigg\vert_{-2}^3\\ &=-\frac{3^{3}-(-2)^3}{3} - 3 \cdot \frac{3^2-(-2)^2}{2} + 7 (3-(-2))\\ &=-\frac{35}{3} - 3 \cdot \frac{5}{2} + 35\\ &=\boxed{\frac{95}{6}}. \end{align} {/eq}

Learn more about this topic:

The Fundamental Theorem of Calculus

from Math 104: Calculus

Chapter 12 / Lesson 10

Related to this Question

Explore our homework questions and answers library