Evaluate the integrals and combine the fractions.

Question

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to your questions.

1) Evaluate $\int_{-1}^{3} (3x^2 - 2x)\,dx$

Find the indefinite integral of the function.
$\int (3x^2 - 2x)\,dx = 3\left(\frac{x^{2+1}}{2+1}\right) - 2\left(\frac{x^{1+1}}{1+1}\right) + C$ $= 3\left(\frac{x^3}{3}\right) - 2\left(\frac{x^2}{2}\right) + C$ $= x^3 - x^2 + C$
Apply the limits of integration using the Fundamental Theorem of Calculus, $F(b) - F(a)$ .
$\left[x^3 - x^2\right]_{-1}^{3} = ((3)^3 - (3)^2) - ((-1)^3 - (-1)^2)$ $= (27 - 9) - (-1 - 1)$ $= 18 - (-2)$ $= 18 + 2$ $= \boxed{20}$
2) Express $\frac{9}{1-4x} - \frac{8}{1-3x}$ as a single fraction.
Find a common denominator for the two fractions, which is $(1-4x)(1-3x)$ .
Rewrite each fraction with the common denominator.
$\frac{9}{1-4x} - \frac{8}{1-3x} = \frac{9(1-3x)}{(1-4x)(1-3x)} - \frac{8(1-4x)}{(1-4x)(1-3x)}$
Combine the numerators over the common denominator.
$= \frac{9(1-3x) - 8(1-4x)}{(1-4x)(1-3x)}$
Expand and simplify the numerator.
$= \frac{9 - 27x - (8 - 32x)}{(1-4x)(1-3x)}$ $= \frac{9 - 27x - 8 + 32x}{(1-4x)(1-3x)}$ $= \frac{(9 - 8) + (-27x + 32x)}{(1-4x)(1-3x)}$ $= \boxed{\frac{1 + 5x}{(1-4x)(1-3x)}}$

Step-by-step solution