Solution: Indefinite Integrals #1

Solution: Indefinite Integrals #1

Evaluate:

[latex]\Large{\int(2x^2+3x-1)dx}[/latex]

Step 1:

 

We can first apply the integration to each term as follows:

 

[latex]\large{\int(2x^2+3x-1)dx = \int(2x^2)dx + \int(3x)dx – \int(1)dx}[/latex]

 

Step 2:

 

Now, we can use the integral power rule to evaluate each integral individually, as shown:

 

[latex]\large{\int(2x^2)dx = \frac{2}{3}x^3}[/latex]

 

[latex]\large{\int(3x)dx = \frac{3}{2}x^2}[/latex]

 

[latex]\large{\int(1)dx = x}[/latex]

 

Step 3:

 

Now, combining everything, we have:

 

[latex]\large{\int(2x^2+3x-1)dx = \frac{2}{3}x^3 + \frac{3}{2}x^2 – x + C}[/latex]

 

Solution Complete!

 

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Solution: Indefinite Integrals #1

Evaluate:

[latex]\Large{\int(2x^2+3x-1)dx}[/latex]

Step 1:

 

We can first apply the integration to each term as follows:

 

[latex]\large{\int(2x^2+3x-1)dx}[/latex]

 

[latex]\large{ = \int(2x^2)dx + \int(3x)dx – \int(1)dx}[/latex]

 

Step 2:

 

Now, we can use the integral power rule to evaluate each integral individually, as shown:

 

[latex]\large{\int(2x^2)dx = \frac{2}{3}x^3}[/latex]

 

[latex]\large{\int(3x)dx = \frac{3}{2}x^2}[/latex]

 

[latex]\large{\int(1)dx = x}[/latex]

 

Step 3:

 

Now, combining everything, we have:

 

[latex]\small{\int(2x^2+3x-1)dx = \frac{2}{3}x^3 + \frac{3}{2}x^2 – x + C}[/latex]

 

Solution Complete!

 

Send us a review!

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