Evaluate the definite integral. ∫02x2−2x+1dx

Question

Evaluate the definite integral.  ∫02x2−2x+1dx

Julie Mathew · Accepted Answer

Step 1
Using the following integrations to solve:

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

\int \frac{1}{x} d x = \ln | x | + C

where C is the constant when indefinite integral is there, but in question we have definite integral so after solving the integration, given limits have to be substituted.
Step 2
Solving the integration as,

\int_{0}^{2} \frac{x^{2} - 2}{x + 1} d x = \int_{0}^{2} ((x - 1) - (\frac{1}{x + 1})) d x

Now integrating it using the formulas,

\int_{0}^{2} \frac{x^{2} - 2}{x + 1} d x = {[\frac{x^{2}}{2} - x - \ln | x + 1 |]}_{0}^{2} = \frac{1}{2} (4) - 2 - \ln | 3 |

= 2 - 2 - \ln | 3 |

= - \ln | 3 |

Troy Lesure · Accepted Answer

Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:

\int_{a}^{b} f (x) d x = F (x) ∣_{a}^{b} = F (b) - F (a)

Step 2: In this case,

f (x) = \frac{x^{2} - 2}{x + 1}

. Find its integral.

\frac{x^{2}}{2} - x - \ln (x + 1) ∣_{0}^{2}

Step 3: Since

F (x) ∣_{a}^{b} = F (b) - F (a)

, expand the above into F(2)−F(0):

(\frac{2^{2}}{2} - 2 - \ln (2 + 1)) - (\frac{0^{2}}{2} - 0 - \ln (0 + 1))

Step 4: Simplify.

- \ln 3

Evaluate the definite integral. \int_{0}^{2}\frac{x^{2}-2}{x+1}dx

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