Background of Definite Integrals
Definite Integrals are calculated by subtracting the function values of upper and lower limits.
$$\displaystyle \int_{a}^{b}{f(x)}dx = F(a) – F(b), \text{where } \frac{d}{dx}F(x) = f(x)$$
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 \( \displaystyle \int_{2}^{3}{(2x+k)}dx = 9 \).
\( \begin{aligned} \displaystyle
\bigg[\frac{2x^{1+1}}{1+1} + kx\bigg]_{2}^{3} &= 9 \\
\big[x^2 + kx\big]_{2}^{3} &= 9 \\
(3^2 + 3k)-(2^2+2k) &= 9 \\
5+k &= 9 \\
\therefore k &= 4 \\
\end{aligned} \\ \)
