Pascal, 115 B (non‐competing)^†

Declaring and defining your own function in Pascal introduces a ton of overhead; it is shorter to expand the integral: $$ \begin{align} \int_{d}^{e} \! \left( a \, x^{2} + b \, x + c \right) \mathrm{d}x =& \int_{d}^{e} \! \left( a \, x^{2} \right) \mathrm{d}x + \int_{d}^{e} \! \left( b \, x^{1} \right) \mathrm{d}x + \int_{d}^{e} \! \left( c \, x^{0} \right) \mathrm{d}x \\ =&\; a \int_{d}^{e} \! \left(x^{2} \right) \mathrm{d}x + b \int_{d}^{e} \! \left(x^{1} \right) \mathrm{d}x + c \int_{d}^{e} \! \left(x^{0} \right) \mathrm{d}x \\ =&\; a \left( \left. \frac{x^3}{3} + p \right) \right|_{d}^{e} + b \left( \left. \frac{x^2}{2} + q \right) \right|_{d}^{e} + c \left( \left. \frac{x^1}{1} + r \right) \right|_{d}^{e} \\ = &\;a \left( \left(\frac{e^3}{3} + p \right) - \left(\frac{d^3}{3} + p \right) \right) + \\ &\;b \left( \left(\frac{e^2}{2} + q \right) - \left(\frac{d^2}{2} + q \right) \right) + \\ &\;c \left( \left(\frac{e^1}{1} + r \right) - \left(\frac{d^1}{1} + r \right) \right) \\ = &\;a \left( \frac{e^3}{3} + p - \frac{d^3}{3} - p \right) + \\ &\;b \left( \frac{e^2}{2} + q - \frac{d^2}{2} - q \right) + \\ &\;c \left( \frac{e^1}{1} + r - \frac{d^1}{1} - r \right) \\ =&\; a \left( \frac{e^3}{3} - \frac{d^3}{3} \right) + b \left( \frac{e^2}{2} - \frac{d^2}{2} \right) + c \left( \frac{e^1}{1} - \frac{d^1}{1} \right) \\ =&\; a \left( \frac{e^3 - d^{3}}{3} \right) + b \left( \frac{e^2 - d^{2}}{2} \right) + c \left( \frac{e^1 - d^{1}}{1} \right) \\ =&\; \frac{a \left( e^3 - d^{3} \right) }{3} + \frac{b \left( e^2 - d^{2} \right) }{2} + c \left( e - d \right) \end{align} $$

program p(input,output);var a,b,c,d,e:real;begin
read(a,b,c,d,e);write(a*(e*e*e-d*d*d)/3+b*(e*e-d*d)/2+c*(e-d))end.

Pretty‐printed:

program areaUnderACurve(input, output);
    var
        a, b, c, d, e: real;
    begin
        readLn(a, b, c, d, e);
        writeLn(a * (e * e * e - d * d * d) / 3 +
                b * (e * e     - d * d    ) / 2 +
                c * (e         - d        )     )
    end.

^†Pascal’s built‐in data type real provides only an implementation‐defined “reasonable” approximation of the actual mathematically correct and precise result, hence it is impossible to establish a precision of ±10⁻³. In practice values of large magnitude lead to errors exceeding the required grade of precision.

JavaScript, 52 bytes

(a,b,c,d,e)=>a*e^3/3+b*e^2/2+c*e-a*d^3/3-b*d^2/2-c*d

Calculates the indefinite integral at e and d, then subtract.

CJam, 27 bytes

q~W%2/~~[2*\~3*\6*0]fb:-6d/

Example input:

[1 2 3 4 8]

Output:

209.33333333333334

CJam, 46 43 bytes

l~\:D-A3#:K*),])\f{\dK/D+\~\Z$*+@_*@*+K/}:+

The arguments are read via STDIN like

1 2 3 4 7

and the output is printed to STDOUT like

135.04650050000018

You can increase the precision by changing the 3 in A3#:K to a higher number. The higher the number, the longer the program will take to finish.

Try it online here

R, 71 bytes

If the input is a vector x <- c(a, b, c, d, e) then you can compute the integral directly as per @feersum's comment.

e=x[5];d=x[4];x[1]*e^3/3+x[2]*e^2/2+x[3]*e-x[1]*d^3/3-x[2]*d^2/2-x[3]*d