First, set up the integral expression:

$\displaystyle \int 2x+1dx$

Now, integrate. Remember to raise the exponent by 1 and then put that result on the denominator:

$\displaystyle \frac{2x^2}{2}+x=x^2+x+C$

Plug in your initial conditions to find C:

$\displaystyle 1^2+1+C=3; C=1$

Now plug back in to get your initial f(x) function:

$\displaystyle f(x)=x^2+x+1$

$\displaystyle \int_{a}^{b} f(x) dx$

In this case, $\displaystyle f(x) = \frac{1}{2} x$.

The antiderivative is  $\displaystyle F(x) = \frac{1}{4} x^2$.

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{1}^{2} \frac{1}{2} x \ dx = \left (\frac{1}{4} x^2 \right )_{1}^{2}$

$\displaystyle = \frac{1}{4} (2)^2 - \left ( \frac{1}{4} (1)^2 \right )$

$\displaystyle = \frac{3}{4}$

$\displaystyle \int_{a}^{b} f(x) dx$

In this case, $\displaystyle f(x) = e^ {x -7}$.

The antiderivative is  $\displaystyle F(x) = e^ {x -7}$.

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{12}^{14} e^ {x -7}\ dx = \left (e^ {x -7} \right )_{12}^{14}$

$\displaystyle = e^ {14 -7} - \left ( e^ {12 -7} \right )$

$\displaystyle = 948.2$

$\displaystyle \int_{a}^{b} f(x) dx$

In this case, $\displaystyle f(x) =\frac{x-4}{x-5}$.

The antiderivative is  $\displaystyle F(x) =\ln(x-5) + x$.

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{6}^{7} \frac{x-4}{x-5} \ dx = \left (\ln(x-5) + x \right )_{6}^{7}$

$\displaystyle = \ln(7-5) + 7 - \left ( \ln(6-5) + 6 \right )$

$\displaystyle = 13.7$

$\displaystyle \int_{a}^{b} f(x) dx$

In this case, $\displaystyle f(x) = \cos(2x-9)$.

The antiderivative is  $\displaystyle F(x) = \frac{\sin(2x-9)}{2}$.

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{0}^{1} \cos(2x-9) \ dx = \left ( \frac{\sin(2x-9)}{2} \right )_{0}^{1}$

$\displaystyle = \frac{\sin(2(1)-9)}{2} - \left ( \frac{\sin(2(0)-9)}{2} \right )$

$\displaystyle = 0.02$

$\displaystyle \int_{a}^{b} f(x) dx$

In this case, $\displaystyle f(x) = -\frac{\sin(2x)}{2}$.

The antiderivative is  $\displaystyle F(x) = \frac{\cos(2x)}{4}$.

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{1}^{2} -\frac{\sin(2x)}{2} \ dx= \left ( \frac{\cos(2x)}{4} \right )_{1}^{2}$

$\displaystyle = \frac{\cos(2(2))}{4} - \left ( \frac{\cos(2(1))}{4} \right )$

$\displaystyle = -0.06$

$\displaystyle \int_{a}^{b} f(x) dx$

In this case, $\displaystyle f(x) =\sec^2 (12x-1)$.

The antiderivative is  $\displaystyle F(x) = \frac{\tan(12x-1)}{12}$.

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{2}^{3}\sec^2 (12x-1)\ dx = \left ( \frac{\tan(12x-1)}{12} \right )_{2}^{3}$

$\displaystyle = \frac{\tan(12(3)-1)}{12} - \left ( \frac{\tan(12(2)-1)}{12} \right )$

$\displaystyle = 0.02$

First, integrate this expression. Remember to raise the exponent by 1 and then put that result on the denominator:

$\displaystyle \frac{x^4}{4}+\frac{4x^3}{3}-\frac{x^2}{2}+x$

Now, evaluate at 2 and then 0. Subtract the results:
$\displaystyle (\frac{16}{4}+\frac{32}{3}-\frac{4}{2}+2)-0=4+\frac{32}{3}-2+2=\frac{44}{3}$

First, integrate this expression. Remember to add one to the exponent and then also put that result on the denominator:

$\displaystyle \frac{x^{\frac{7}{4}}}{\frac{7}{4}}$

Simplify:

$\displaystyle \frac{4}{7}(x^{\frac{7}{4}})$

Evaluate at 2 and then 1. Subtract the results:

$\displaystyle (\frac{4}{7}(2^{\frac{7}{4}}))-(\frac{4}{7}(1^{\frac{7}{4}}))$

Round to four places:

$\displaystyle 1.9220-.5714=1.3506$

First, chop up the fraction into two separate terms:

$\displaystyle \int \frac{5}{x}+xdx$

Now, integrate:

$\displaystyle 5ln\left | x \right |+\frac{x^2}{2}$

Evaluate at 4 and then at 1. Subtract the results:

$\displaystyle (5ln4+8)-(5ln1+\frac{1}{2})$

Round to four places:
$\displaystyle 14.9315-.5=14.4315$