Q. 1 - Calculus

Question

Calculate each of the following definite integrals, using integration techniques and fundamental theorem of calculus

$$\begin{gathered} {\int }_{-3}^3(9-x^2)dx \\ {\int }_0^1(x^4+2x^2)dx \\ {\int }_0^1\left(4\right)dx \\ {\int }_{-3}^3(4-9{\left(\ln x\right)}^2)dx \\ {\int }_0^2(4y^2-y^4)dy \\ {\int }_0^1(2-\sqrt{y})dy \end{gathered}$$

Step-by-Step Solution

Answer

We have

$$\begin{gathered} {\int }_{-3}^3(9-x^2)dx=36 \\ {\int }_0^1(x^4+2x^2)dx=\frac{13}{15} \\ {\int }_0^1\left(4\right)dx=4 \\ {\int }_{-3}^3(4-9{\left(\ln x\right)}^2)dx=-\frac{4}{3}{e}^{\frac{2}{3}}-6 \\ {\int }_0^2(4y^2-y^4)dy=\frac{64}{15} \\ {\int }_0^1(2-\sqrt{y})dy=\frac{29}{6} \end{gathered}$$

Step 1: Given information

We are given definite integrals we have to evaluate them.

Part a) Step 1: Evaluate

We get,

$$\begin{gathered} {\int }_{-3}^3(9-x^2)dx \\ ={{[9x-\frac{x^3}{3}]}^3}_{-3}=[27-\frac{27}{3}]-[-27-(-\frac{27}{3}\left)\right] \\ =54-18 \\ =36 \end{gathered}$$

Part b) Step 1: Evaluate

We have,

$$\begin{gathered} {\int }_0^1(x^4+2x^2)dx \\ =\frac{x^5}{5}+\frac{2x^3}{3} \\ =\frac{1}{5}+\frac{2}{3} \\ =\frac{13}{15} \end{gathered}$$

Part c) Step 1: Evaluate

We have,

$$\begin{gathered} {\int }_0^{1}4dx \\ ={{\left[4x\right]}^1}_0=4 \end{gathered}$$

Part d) Step 1: Evaluate

We have,

$$\begin{gathered} {\int }_1^{e^{\frac{2}{3}}}(4-9{\left(\ln x\right)}^2)dx \\ Put \ln x =y \\ hence \frac{1}{x}dx=dy \\ therefore dx=e^{y}dy \\ Hence the integral becomes \\ {\int }_0^{\frac{2}{3}}(4-9y^2)e^{y}dy \\ ={\int }_0^{\frac{2}{3}}4e^y-9y^2{e}^{y}dy \\ ={{[4e^y-9y^2{e}^y-2y{e}^y+2e^y]}^{\frac{2}{3}}}_0=-\frac{4}{3}{e}^{\frac{2}{3}}-6 \end{gathered}$$

Step e) Step 1: Evaluate

We have,

$$\begin{gathered} {\int }_0^2(4y^2-y^4)dy \\ ={{[\frac{4}{3}{y}^3-\frac{y^5}{5}]}^2}_0=\frac{32}{3}-\frac{32}{5} \\ =\frac{64}{15} \end{gathered}$$

Part f) Step 1: Evaluate

We have,

$$\begin{gathered} {\int }_0^1{(2-\sqrt{y})}^{2}dy \\ ={\int }_0^1(4-4\sqrt{y}+y)dy \\ ={{[4y+\frac{2}{3}{y}^{\frac{2}{3}}+\frac{y^2}{2}]}_0}^1 \\ =4+\frac{2}{3}+\frac{1}{2} \\ =\frac{29}{6} \end{gathered}$$