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8.4. Integration by substitution [P3]

8.4. Integration by substitution [P3]

Assessment

Presentation

Mathematics

12th Grade

Practice Problem

Medium

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Math Math

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16 Slides • 13 Questions

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Math Response

Expand the power.

(x+1)2dx=??? dx\int_{ }^{ }\left(x+1\right)^2dx=\int_{ }^{ }???\ dx

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Math Response

Evaluate the integral.

(x+1)2dx=x2+2x+1 dx=???\int_{ }^{ }\left(x+1\right)^2dx=\int_{ }^{ }x^2+2x+1\ dx=???

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Consider!

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8.4. Integration by substitution

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Learning Objectives

At the end of the lesson, learners will be able to:

  • Use a given substitution to simplify and evaluate integral.

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Substitution method and Chain rule

Integration by substitution can be considered as the reverse process of differentiation by chain rule.

This method is used when a simple substitution can be applied that will transform a complicated integral into a simpler integral.

The integral must be completely rewritten in terms of the new variable.





In Mathematics Pure 3, the substitution will be given.

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Substitution method

In the following example, we will see how we can use substitution method to simplify integration.

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Substitution method

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Math Response

Substitute t=x+1t=x+1   to integrate (x+1)7dx\int_{ }^{ }\left(x+1\right)^7dx   .
Since t=x+1t=x+1   , then (x+1)7=???\left(x+1\right)^7=???   .

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11

Math Response

Substitute t=x+1t=x+1   to integrate (x+1)7dx\int_{ }^{ }\left(x+1\right)^7dx   .
Since t=x+1t=x+1   , then (x+1)7=t7\left(x+1\right)^7=t^7   .

So, (x+1)7dx=???dx\int_{ }^{ }\left(x+1\right)^7dx=\int_{ }^{ }???dx

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Substitution method

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Math Response

Since t=x+1t=x+1   , when we differentiate we have

dtdx=???\frac{dt}{dx}=???

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Substitution method

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Substitution method

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Math Response

Since t=x+1t=x+1   , then (x+1)7=t7\left(x+1\right)^7=t^7   .
and also dtdx=1\frac{dt}{dx}=1   , so dx=dtdx=dt   .

So, (x+1)7dx=t7dt=???\int_{ }^{ }\left(x+1\right)^7dx=\int_{ }^{ }t^7dt=??? .

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17

Math Response

Since t=x+1t=x+1   , then (x+1)7=t7\left(x+1\right)^7=t^7   .
and also dtdx=1\frac{dt}{dx}=1   , so dx=dtdx=dt   .

So, (x+1)7dx=t7dt=18t8+c=???\int_{ }^{ }\left(x+1\right)^7dx=\int_{ }^{ }t^7dt=\frac{1}{8}t^8+c=??? .

(the original question is in variable "x", so the answer should also be in "x")

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Substitution method

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Substitution method

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Math Response

Since t=x23t=x^2-3 , then:

dtdx=???\frac{dt}{dx}=???

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24

Math Response

Since t=x23t=x^2-3 , then dtdx=2x\frac{dt}{dx}=2x .

Thus,

dx=??? dtdx=???\ dt

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25

Math Response

Since t=x23t=x^2-3 , then dtdx=2x\frac{dt}{dx}=2x .

Thus, dx=12x dtdx=\frac{1}{2x}\ dt .

So, the integral become:

x x23dx=x ???dx\int_{ }^{ }\frac{x\ }{\sqrt[]{x^2-3}}dx=\int_{ }^{ }\frac{x\ }{\sqrt[]{???}}dx

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26

Math Response

Since t=x23t=x^2-3 , then dtdx=2x\frac{dt}{dx}=2x .

Thus, dx=12x dtdx=\frac{1}{2x}\ dt .

So, the integral become:

x x23dx=x t1 2xdt=???dt\int_{ }^{ }\frac{x\ }{\sqrt[]{x^2-3}}dx=\int_{ }^{ }\frac{x\ }{\sqrt[]{t}}\frac{1\ }{2x}dt=\int_{ }^{ }???dt

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27

Math Response

Since t=x23t=x^2-3 , then dtdx=2x\frac{dt}{dx}=2x .

Thus, dx=12x dtdx=\frac{1}{2x}\ dt .

So, the integral become:

x x23dx=x t1 2xdt=1 2tdt=1 2t12dt=???\int_{ }^{ }\frac{x\ }{\sqrt[]{x^2-3}}dx=\int_{ }^{ }\frac{x\ }{\sqrt[]{t}}\frac{1\ }{2x}dt=\int_{ }^{ }\frac{1\ }{2\sqrt[]{t}}dt=\int_{ }^{ }\frac{1\ }{2}t^{-\frac{1}{2}}dt=???

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28

Math Response

Since t=x23t=x^2-3 , then dtdx=2x\frac{dt}{dx}=2x .

Thus, dx=12x dtdx=\frac{1}{2x}\ dt .

So, the integral become:

x x23dx=x t1 2xdt=\int_{ }^{ }\frac{x\ }{\sqrt[]{x^2-3}}dx=\int_{ }^{ }\frac{x\ }{\sqrt[]{t}}\frac{1\ }{2x}dt= 1 2tdt=1 2t12dt=t12+c=???\int_{ }^{ }\frac{1\ }{2\sqrt[]{t}}dt=\int_{ }^{ }\frac{1\ }{2}t^{-\frac{1}{2}}dt=t^{\frac{1}{2}}+c=???

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