Definite Integrals Practice Problems

Problem: Definite Integrals #1

Evaluate:

$\Large{\int_{0}^{2}(3x^2-2x+1)dx}$

Solution

Step 1:

Using our knowledge of the integral power rule, we can find the integral as follows:

\large{\int_{0}^{2}(3x^2-2x+1) dx = (x^3-x^2+x)|_{0}^{2}}

Step 2:

Now, we can apply the fundamental theorem of calculus. This states that:

\large{\int_{a}^{b}f(x) dx = F(b) – F(a)}

Thus, we have:

\large{\int_{0}^{2}(3x^2-2x+1) dx = [(2)^3 – (2)^2 + 2] – [(0)^3-(0)^2+0 ] }

\large{ = 6 \:-\: 0 = 6}

Step 3:

So, our final answer is:

\large{\int_{0}^{2}(3x^2-2x+1)dx = 6}

Solution Complete!

Solution

Step 1:

Using our knowledge of the integral power rule, we can find the integral as follows:

\large{\int_{0}^{2}(3x^2-2x+1) dx =}

\large{ (x^3-x^2+x)|_{0}^{2}}

Step 2:

Now, we can apply the fundamental theorem of calculus. This states that:

\large{\int_{a}^{b}f(x) dx = F(b) – F(a)}

Thus, we have:

\large{\int_{0}^{2}(3x^2-2x+1) dx =}

\large{ [(2)^3 – (2)^2 + 2] – [(0)^3-(0)^2+0 ] }

\large{ = 6 \:-\: 0 = 6}

Step 3:

So, our final answer is:

\large{\int_{0}^{2}(3x^2-2x+1)dx = 6}

Solution Complete!

Problem: Definite Integrals #2

Evaluate:

$\Large{\int_{0}^{\frac{\pi}{2}}sin(x)dx}$

Problem: Definite Integrals #3

Evaluate:

$\Large{\int_{1}^{3}(cos(x) + \frac{2}{x^4})dx}$

Problem: Definite Integrals #4

Evaluate:

$\Large{\int_{3}^{1}(12t^4 - 3t^3 + 4t - 3)dt}$

Need Additional Help? Chat with a tutor now!

QUESTIONS?

If you have any questions about our services or have any feedback, do not hesitate to get in touch with us!