Definite Integrals Practice Problems
Problem: Definite Integrals #1
Evaluate:
[latex]\Large{\int_{0}^{2}(3x^2-2x+1)dx}[/latex]
Step 1:
Using our knowledge of the integral power rule, we can find the integral as follows:
[latex]\large{\int_{0}^{2}(3x^2-2x+1) dx = (x^3-x^2+x)|_{0}^{2}}[/latex]
Step 2:
Now, we can apply the fundamental theorem of calculus. This states that:
[latex]\large{\int_{a}^{b}f(x) dx = F(b) – F(a)}[/latex]
Thus, we have:
[latex]\large{\int_{0}^{2}(3x^2-2x+1) dx = [(2)^3 – (2)^2 + 2] – [(0)^3-(0)^2+0 ] }[/latex]
[latex]\large{ = 6 \:-\: 0 = 6}[/latex]
Step 3:
So, our final answer is:
[latex]\large{\int_{0}^{2}(3x^2-2x+1)dx = 6}[/latex]
Solution Complete!
Step 1:
Using our knowledge of the integral power rule, we can find the integral as follows:
[latex]\large{\int_{0}^{2}(3x^2-2x+1) dx =}[/latex]
[latex]\large{ (x^3-x^2+x)|_{0}^{2}}[/latex]
Step 2:
Now, we can apply the fundamental theorem of calculus. This states that:
[latex]\large{\int_{a}^{b}f(x) dx = F(b) – F(a)}[/latex]
Thus, we have:
[latex]\large{\int_{0}^{2}(3x^2-2x+1) dx =}[/latex]
[latex]\large{ [(2)^3 – (2)^2 + 2] – [(0)^3-(0)^2+0 ] }[/latex]
[latex]\large{ = 6 \:-\: 0 = 6}[/latex]
Step 3:
So, our final answer is:
[latex]\large{\int_{0}^{2}(3x^2-2x+1)dx = 6}[/latex]
Solution Complete!
Problem: Definite Integrals #2
Evaluate:
[latex]\Large{\int_{0}^{\frac{\pi}{2}}sin(x)dx}[/latex]
[latex]\Large{\int_{0}^{\frac{\pi}{2}}sin(x)dx}[/latex]
Problem: Definite Integrals #3
Evaluate:
[latex]\Large{\int_{1}^{3}(cos(x) + \frac{2}{x^4})dx}[/latex]
[latex]\Large{\int_{1}^{3}(cos(x) + \frac{2}{x^4})dx}[/latex]
Problem: Definite Integrals #4
Evaluate:
[latex]\Large{\int_{3}^{1}(12t^4 - 3t^3 + 4t - 3)dt}[/latex]
[latex]\Large{\int_{3}^{1}(12t^4 - 3t^3 + 4t - 3)dt}[/latex]
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