Evaluating Integrals with Partial Fractions

Evaluating Integrals with Partial Fractions

Assessment

Interactive Video

Mathematics

9th - 10th Grade

Hard

Created by

Thomas White

FREE Resource

The video tutorial explains how to use partial fractions to solve integrals where the degree of the polynomial in the numerator is lower than that in the denominator. It covers factorizing the denominator, setting up the partial fraction equation, solving for constants, and evaluating the integral.

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15 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the relationship between the degrees of the numerator and denominator in the given integral?

The numerator has a higher degree.

The degrees are equal.

The numerator has a lower degree.

The denominator is a constant.

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which method is suggested for solving the integral when the numerator's degree is lower than the denominator's?

Trigonometric substitution

Partial fractions

Substitution

Integration by parts

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What are the factors obtained after factorizing the denominator?

x + 2 and x - 3

x - 2 and x + 3

x + 2 and x + 3

x - 2 and x - 3

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How are the partial fractions expressed with distinct real factors?

As a single fraction

As a sum of constants

As a product of fractions

As a sum of fractions with constants in the numerators

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the first step in solving for the constants in the partial fraction decomposition?

Add a constant to both sides

Integrate both sides

Multiply through by the denominator

Differentiate both sides

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the result of multiplying through by the denominator in the partial fraction setup?

A logarithmic equation

A differential equation

A trigonometric equation

A polynomial equation

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the significance of having distinct real factors in the denominator?

It allows for a unique solution

It simplifies the polynomial

It complicates the integration

It has no significance

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