# Definite Integrals – Definition and Examples

## Background of Definite Integrals

Definite Integrals are calculated by subtracting the function values of upper and lower limits.
$$\large \displaystyle \int_{a}^{b}{f(x)}dx = F(a)-F(b), \text{where } \frac{d}{dx}F(x) = f(x)$$

# Practice Questions

### Question 1

Evaluate $\displaystyle \int_{0}^{2}{6x}dx$.

\begin{aligned} \displaystyle \int_{0}^{2}{6x}dx &= \bigg[\frac{6x^{1+1}}{1+1}\bigg]_{0}^{2} \\ &= \big[3x^2\big]_{0}^{2} \\ &= 3 \times 2^2-3 \times 0^2 \\ &= 12 \end{aligned}

### Question 2

Evaluate $\displaystyle \int_{2}^{4}5dx$.

\begin{aligned} \displaystyle \int_{2}^{4}5dx &= \big[5x\big]_{2}^{4} \\ &= 5 \times 4-5 \times 2 \\ &= 10 \end{aligned}

### Question 3

Evaluate $\displaystyle \int_{1}^{3}{(2x+4)}dx$.

\begin{aligned} \displaystyle \int_{1}^{3}{(2x+4)}dx &= \bigg[\frac{2x^{1+1}}{1+1} + 4x\bigg]_{1}^{3} \\ &= \big[x^2 + 4x\big]_{1}^{3} \\ &= (3^2 + 4 \times 3)-(1^2 + 4 \times 1) \\ &= 16 \end{aligned}

### Question 4

Evaluate $\displaystyle \int_{0}^{1}{(2x-1)^3}dx$.

\begin{aligned} \displaystyle \int_{0}^{1}{(2x-1)^3}dx &= \bigg[\dfrac{(2x-1)^{3+1}}{2(3+1)}\bigg]_{0}^{1} \\ &= \frac{1}{8}\Big[(2x-1)^4\Big]_{0}^{1} \\ &= \frac{1}{8}\Big[(2 \times 1-1)^4-(2 \times 0-1)^4\Big] \\ &= 0 \end{aligned}

### Question 5

Find $k$, if $\displaystyle \int_{2}^{3}{(2x+k)}dx = 9$.

\begin{aligned} \displaystyle \bigg[\frac{2x^{1+1}}{1+1} + kx\bigg]_{2}^{3} &= 9 \\ \big[x^2 + kx\big]_{2}^{3} &= 9 \\ (3^2 + 3k)-(2^2+2k) &= 9 \\ 5+k &= 9 \\ \therefore k &= 4 \end{aligned}