Definite Integrals – Examples with Answers

To solve this definite integral, we have to start by integrating the expression and use square brackets to indicate the limits of integration:

$$\int_{0}^{2} 4x^3dx=[x^4+c]_{0}^{2}$$

Now, we can evaluate the limits. The expression evaluated at the lower limit is subtracted from the expression evaluated at the upper limit:

$$[x^4+c]_{0}^{2}=[(2)^4+c]-[(0)^4+c]$$

When we simplify this, we have:

$latex =[(2)^4+c]-[(0)^4+c]$

$latex =[16+c]-[0+c]$

$latex =16$

We can see that the constant of integration was removed, so we can omit it when we are working with definite integrals.

We start by finding the integral to be evaluated. Keep the limits of integration using square brackets and ignore the constant of integration:

$$\int_{2}^{3} (6x^2-1)dx=[2x^3-x]_{2}^{3}$$

Now that we have the integral, we can evaluate the given limits as follows:

$$[2x^3-x]_{2}^{3}=[2(3)^3-(3)]-[2(2)^3-(2)]$$

Finally, we simplify as follows:

$latex =[2(3)^3-(3)]-[2(2)^3-(2)]$

$latex =[54-3]-[16-2]$

$latex =[51]-[14]$

$latex =37$

Integrating the given expression and maintaining the limits of integration, we have:

$$\int_{4}^{5} (4x+3)dx=[2x^2+3x]_{4}^{5}$$

Evaluating the limits, we have:

$$[2x^2+3x]_{4}^{5}=[2(5)^2+3(5)]-[2(4)^2+3(4)]$$

Finally, we simplify to obtain:

$latex =[2(5)^2+3(5)]-[2(4)^2+3(4)]$

$latex =[50+15]-[32+12]$

$latex =[65]-[44]$

$latex =21$

To find the value of the integral, we have to start by integrating the expression while keeping the limits of integration:

$$\int_{2}^{3} (4-3x^2)dx=[4x-x^3]_{2}^{3}$$

Now, let’s evaluate the limits. Subtract the lower limit from the upper limit:

$$[4x-x^3]_{2}^{3}=[4(3)-(3)^3]-[4(2)-(2)^3]$$

We can simplify to get a single value:

$latex =[4(3)-(3)^3]-[4(2)-(2)^3]$

$latex =[12-27]-[8-8]$

$latex =[-15]-[0]$

$latex =-15$

We have to start by finding the integral of the given expression. In this case, we are going to use the laws of exponents to write as follows:

$$\int_{2}^{8} \frac{1}{x^2} dx=\int_{2}^{8} x^{-2} dx$$

$$=[-x^{-1}]_{2}^{8}$$

Now that we have the integral, let’s evaluate the limits:

$$[-x^{-1}]_{2}^{8}=[-(8)^{-1}]-[-(2)^{-1}]$$

Finally, we simplify to obtain a defined value:

$latex =[-(8)^{-1}]-[-(2)^{-1}]$

$$=-\frac{1}{8}+\frac{1}{2}$$

$$=\frac{3}{8}$$

To integrate this expression, we have to use the laws of exponents to write it without the fraction. Then, we have:

$$\int_{1}^{2} \frac{4}{x^3} dx=\int_{1}^{2} 4x^{-3} dx$$

$$=[-2x^{-2}]_{1}^{2}$$

Now, we are going to evaluate the given expression with the given limits of integration:

$$[-2x^{-2}]_{1}^{2}=[-2(2)^{-2}]-[-2(1)^{-2}]$$

When we simplify this, we have:

$latex =[-2(2)^{-2}]-[-2(1)^{-2}]$

$$=-\frac{1}{2}+2$$

$$=\frac{3}{2}$$

In this case, we have a square root. Therefore, we write it with a numerical exponent using the laws of exponents:

$$\int_{4}^{9} \sqrt{x} dx=\int_{4}^{9} x^{\frac{1}{2}} dx$$

$$=\left[\frac{2}{3} x^{\frac{2}{3}}\right]_{4}^{9}$$

Now, we evaluate the expression using the limits of integration:

$$=\left[\frac{2}{3} (9)^{\frac{3}{2}}\right]-\left[\frac{2}{3} (4)^{\frac{3}{2}}\right]$$

We can simplify by considering that the exponent $latex \frac{3}{2}$ is equivalent to taking the square root and cubing the result.

$$=\left[\frac{2}{3} (9)^{\frac{3}{2}}\right]-\left[\frac{2}{3} (4)^{\frac{3}{2}}\right]$$

$$=\left[\frac{2}{3} (27)\right]-\left[\frac{2}{3} (8)\right]$$

$$=18-\frac{16}{3}$$

$$=\frac{38}{3}=12~\frac{2}{3}$$

We start by writing the expression with a numerical exponent to find its integral:

$$\int_{1}^{4} \left( 3- \frac{1}{\sqrt{x}}\right)dx=\int_{1}^{4} ( 3- x^{-\frac{1}{2}})dx$$

$$=\left[ 3x-2 x^{\frac{1}{2}}\right]_{0}^{2}$$

Evaluating the expression using the limits of integration, we have:

$$=\left[ 3(4)-2 (4)^{\frac{1}{2}}\right]-\left[ 3(1)-2 (1)^{\frac{1}{2}}\right]$$

Finally, we can simplify considering that the exponent $latex \frac{1}{2}$ is equivalent to the square root:

$$=\left[ 3(4)-2 (4)^{\frac{1}{2}}\right]-\left[ 3(1)-2 (1)^{\frac{1}{2}}\right]$$

$$=[ 12-4]-\left[ 3-2\right]$$

$latex =8-1$

$latex =7$

We rewrite the exponents using the laws of exponents and find the integral of the expression:

$$\int_{\frac{1}{2}}^{1} 1+\frac{1}{x^2} dx=\int_{\frac{1}{2}}^{1} 1+x^{-2} dx$$

$$=\left[x-x^{-1} \right]_{\frac{1}{2}}^{1}$$

Now, we evaluate the expression obtained using the limits of integration:

$$=\left[1-(1)^{-1} \right]-\left[\frac{1}{2}-(\frac{1}{2})^{-1} \right]$$

By simplifying, we can obtain a single value:

$$=\left[1-1 \right]-\left[\frac{1}{2}-2 \right]$$

$$=-\left[\frac{1}{2}-2 \right]$$

$$=\frac{3}{2}$$

Let’s find the integral of the expression by writing the cube root as a numerical exponent:

$$\int_{1}^{8} \sqrt[3]{x} dx=\int_{1}^{8} x^{\frac{1}{3}} dx$$

$$=\left[\frac{3}{4}x^{\frac{4}{3}} \right]_{1}^{8}$$

Evaluating the limits of integration, we have:

$$=\left[\frac{3}{4}(8)^{\frac{4}{3}} \right]-\left[\frac{3}{4}(1)^{\frac{4}{3}} \right]$$

We can simplify this by considering the exponent $latex \frac{4}{3}$ to be equivalent to taking the cube root of the number and raising the result to the fourth power:

$$=\left[\frac{3}{4}(16) \right]-\left[\frac{3}{4}(1) \right]$$

$$=12-\frac{3}{4}$$

$$=\frac{45}{4}=11~\frac{1}{4}$$