In this case, \(\displaystyle f(x) = \frac{1}{6x+5}\).

The antiderivative is  \(\displaystyle F(x) = \frac{\ln(6x+5)}{6}\).

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

\(\displaystyle \int_{0}^{1} \frac{1}{6x+5} \ dx = \left ( \frac{\ln(6x+5)}{6} \right )_{0}^{1}\)

\(\displaystyle = \frac{\ln(6(1)+5)}{6} - \left ( \frac{\ln(6(0)+5)}{6} \right )\)

\(\displaystyle = 0.13\)

In this case, \(\displaystyle f(x) = 3x - 5\).

The antiderivative is  \(\displaystyle F(x) = \frac{3x^2}{2}-5x\).

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

\(\displaystyle \int_{0}^{1} 3x - 5 \ dx = \left ( \frac{3x^2}{2}-5x \right )_{0}^{1}\)

\(\displaystyle = \frac{3(1)^2}{2}-5(1) - \left ( \frac{3(0)^2}{2}-5(0) \right )\)

\(\displaystyle =-3.5\)

In this case, \(\displaystyle f(x) = e^{8x} + \cos(4x)\).

The antiderivative is  \(\displaystyle F(x) =\frac{ \sin(4x)}{4} + \frac{e^{8x}}{8}\).

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

\(\displaystyle \int_{0}^{1} e^{8x} + \cos(4x) \ dx = \left ( \frac{ \sin(4x)}{4} + \frac{e^{8x}}{8} \right )_{0}^{1}\)

\(\displaystyle =\frac{ \sin(4(1))}{4} + \frac{e^{8(1)}}{8} - \left ( \frac{ \sin(4(0))}{4} + \frac{e^{8(0)}}{8} \right )\)

\(\displaystyle = 372.6\)

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

In this case, \(\displaystyle f(x) = e^{9 - 2x}\).

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

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

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

\(\displaystyle = -\frac{ e^{9 - 2(1)}}{2} - \left ( -\frac{ e^{9 - 2(0)}}{2} \right )\)

\(\displaystyle = 3503.2\)

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

In this case, \(\displaystyle f(x) = x^6 - \frac{1}{3} x^{2}\).

The antiderivative is  \(\displaystyle F(x) =\frac{x^7}{7} - \frac{x^3}{9}\).

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

\(\displaystyle \int_{1}^{2} x^6 - \frac{1}{3} x^{2} \ dx = \left ( \frac{x^7}{7} - \frac{x^3}{9} \right )_{1}^{2}\)

\(\displaystyle = \frac{2^7}{7} - \frac{2^3}{9} - \left ( \frac{1^7}{7} - \frac{1^3}{9} \right )\)

\(\displaystyle = 17.37\)

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

In this case, \(\displaystyle f(x) = 13,525\).

The antiderivative is  \(\displaystyle F(x) = 13,525x\).

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

\(\displaystyle \int_{1}^{2} 13,525 \ dx = \left ( 13,525x \right )_{1}^{2}\)

\(\displaystyle = 13,525(2) - \left ( 13,525(1) \right )\)

\(\displaystyle = 13,525\)

First, integrate the function. Remember, when integrating, raise the exponent by 1 and then put that result on the denominator. The first step should look like this: \(\displaystyle 3(\frac{x^3}{3})-\frac{x^2}{2}+4x=x^3-\frac{x^2}{2}+4x\). Then, evaluate the function at 3 and subtract from the result when you plug in 0. \(\displaystyle (27-\frac{9}{2}+12)-0=39-\frac{9}{2}=\frac{69}{2}\).

To integrate, remember to raise the exponent by 1 and then put that result on the denominator: \(\displaystyle \frac{x^4}{4}-\frac{3x^3}{3}+\frac{x^2}{2}=\frac{x^4}{4}-x^3+\frac{x^2}{2}\). Then, evaluate at 3 and then 1. Subtract the two results. \(\displaystyle (\frac{81}{4}-27+\frac{9}{2})-(\frac{1}{4}-1+\frac{1}{2})=(\frac{99}{4}-27)-(\frac{3}{4}-1)=-2\).

First, integrate the expression, remembering to add one to the exponent and then putting that result on the denominator: \(\displaystyle \frac{x^3}{3}-\frac{3x^2}{2}+x\). Then evaluate at 3 and then 1. Subtract the results: \(\displaystyle (9-\frac{27}{2}+3)-(\frac{1}{3}-\frac{3}{2}+1)=(12-\frac{27}{2})-(-\frac{7}{6}+1)=11-\frac{74}{6}=\frac{-4}{3}\).

\(\displaystyle \int^2_1\frac{3x^4-x^2+7}{x^2}dx=\int^2_1(3x^2-1+7x^{-2})dx=(x^3-x-\frac{7}{x})\left.\begin{matrix} \\ \end{matrix}\right| ^2_1=\frac{19}{2}\)