Question 12 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}$
Answer$\int\limits^\frac{\pi}{4}_0\tan^2\text{x}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_0(\sec^2\text{x}-1)\text{dx}$
$=\big[\tan\text{x}-\text{x}\big]^\frac{\pi}{4}_0$
$=1-\frac{\pi}{4}-0$
$=1-\frac{\pi}{4}$
View full question & answer→Question 22 Marks
Evaluate:
$\int\limits^3_23^\text{x}\text{ dx}$
Answer$\text{I}=\int\limits^3_23^\text{x}$
$=\Big[\frac{3\text{x}}{\log3}\Big]^3_2+\text{C}$ $\Big(\text{Use}:\int\text{a}^{\text{x}}=\frac{\text{a}^{\text{x}}}{\log\text{a}}+\text{C}\Big)$
$=\frac{3^3}{\log3}-\frac{3^2}{\log3}+\text{C}$
$=\frac{1}{\log3}(3^2-3^3)+\text{C}$
$=\frac{1}{\log3}(27-9)+\text{C}$
$=\frac{1}{\log3}(18)+\text{C}$
View full question & answer→Question 32 Marks
Evaluate the following integrals:
$\int\limits^{2}_1\log_\text{e}[\text{x}]\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{2}_1\log_\text{e}[\text{x}]\text{dx}$
We know that,
$[\text{x}]=1,\text{ when }1<\text{x}<2$
$\therefore\ \text{I}=\int\limits^{2}_1\log_\text{e}[\text{x}]\text{dx}$
$\text{I}=\int\limits^{2}_1(0)\text{dx}$
$\text{I}=0$
View full question & answer→Question 42 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{4}_0\tan\text{x dx}$
Answer$\int\limits^\frac{\pi}{4}_0\tan\text{x dx}$
$=\big[\log\sec\text{x}\big]^{\frac{\pi}{4}}_0$
$=\log\sec\frac{\pi}{4}-\log\sec0$
$=\log\sqrt{2}-\log1$
$=\log2^{\frac{1}{2}}-0$
$=\frac{1}{2}\log2$
View full question & answer→Question 52 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\sqrt{1-\cos2\text{x}}\text{ dx}$
Answer$\int\limits^{\frac{\pi}{2}}_0\sqrt{1-\cos2\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\sqrt{2\sin^2\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\sqrt{2}\sin\text{x}\text{ dx}$
$=-\sqrt{2}\Big[\cos\text{x}\Big]^{\frac{\pi}{2}}_0$
$=-\big(0-\sqrt{2}\big)$
$=\sqrt{2}$
View full question & answer→Question 62 Marks
$\int\limits^{1}_0\{\text{x}\}\text{dx},$ where {x} denotes the fractional part of x.
AnswerWe have,
$\text{I}=\int\limits^{1}_0\{\text{x}\}\text{dx}$
We know $\{\text{x}\}=\text{x},0<\text{x}<1$
$\therefore\ \text{I}=\int\limits^{1}_0\{\text{x}\}\text{dx}$
$=\Big[\frac{\text{x}^2}{2}\Big]^1_0$
$=\frac{1}{2}-\frac{0}{2}$
$=\frac{1}{2}$
View full question & answer→Question 72 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\cos\text{x}}\text{ dx}$
AnswerWe use $1+\cos\text{x}=2\cos^2\frac{\text{x}}{2}$
$=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{2\cos^2\frac{\text{x}}{2}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{2}\cos\frac{\text{x}}{2}\text{ dx}$
$=\sqrt{2}\Big[2\sin\frac{\text{x}}{2}\Big]^{\frac{\pi}{2}}_0$
$=2\sqrt{2}\Big[\frac{1}{\sqrt{2}}\Big]$
$=2$
View full question & answer→Question 82 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\infty}\text{e}^{-\text{x}}\text{ dx}$
AnswerWe have
$\int_{0}^\limits{\infty}\text{e}^{-\text{x}}\text{ dx}$
We know that $\int\text{e}^{-\text{x}}=-\text{e}^{-\text{x}}$
$\int_{0}^\limits{\infty}\text{e}^{-\text{x}}\text{ dx}$
$=\big[-\text{e}^{-\text{x}}\big]^{\infty}_0$
$=\big[\text{e}^{-\infty}+\text{e}^{-0}\big]$ $[\because\text{e}^{\infty}=0,\text{ e}^0=1\big]$
$=[-0+1]$
$=1$
View full question & answer→Question 92 Marks
Evaluate the following integrals:
$\int\limits^1_0 \frac{1}{1+\text{x}^2}\text{ dx}$
Answer$\int\limits^1_0 \frac{1}{1+\text{x}^2}\text{ dx}$
$=\big[\tan^{-1}\text{x}\big]^1_0$
$=\frac{\pi}{4}-0$
$=\frac{\pi}{4}$
View full question & answer→Question 102 Marks
If $\int\limits^1_0(3\text{x}^2+2\text{x}+\text{k})\text{dx}=0,$ find the value of k.
AnswerWe have,
$\int\limits^1_0(3\text{x}^2+2\text{x}+\text{k})\text{dx}=0$
$\Rightarrow\big[\text{x}^3+\text{x}^2+\text{kx}\big]^1_0=0$
$\Rightarrow1+1+\text{k}-0=0$
$\Rightarrow\text{k}=-2$
View full question & answer→Question 112 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin^2\text{x}\text{ dx}$
Answer$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin^2\text{x}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\frac{1-\cos2\text{x}}{2}\text{dx}$
$=\frac{1}{2}\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}(1-\cos2\text{x})\text{dx}$
$=\frac{1}{2}\Big[\text{x}-\frac{\sin2\text{x}}{2}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$=\frac{1}{2}\Big(\frac{\pi}{2}-0+\frac{\pi}{2}-0\Big)$
$=\frac{\pi}{2}$
View full question & answer→Question 122 Marks
Evaluate the following integrals:
$\int^\limits{2\pi}_{0}|\sin\text{x}|\text{dx}$
Answer$\int^\limits{2\pi}_{0}|\sin\text{x}|\text{dx}=\int^\limits{\pi}_{0}\sin\text{x }\text{dx}+\int^\limits{2\pi}_{\pi}-\sin\text{x }\text{dx}$
$=\big[-\cos\text{x}\big]^{\pi}_0+\big[\cos\text{x}\big]^{2\pi}_\pi$
$=\big[1+1\big]+\big[1+1\big]$
$\int^\limits{2\pi}_{0}|\sin\text{x}|\text{dx}=4$
View full question & answer→Question 132 Marks
Evaluate the following integrals:
$\int\limits^{1.5}_0\big[\text{x}\big]\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{1.5}_0\big[\text{x}\big]\text{dx}$
$=\int\limits^{1}_0\big[\text{x}\big]\text{dx}+\int\limits^{1.5}_0\big[\text{x}\big]\text{dx}$
$=\int\limits^{1}_0(0)\text{dx}+\int\limits^{1.5}_0(1)\text{dx}$ $\begin{bmatrix}\because\big[\text{x}\big]=\begin{cases}0,&0\leq\text{x}<1\\1,&1\leq\text{x}<1.5\end{cases}\end{bmatrix}$
$=0+\big[\text{x}\big]^{1.5}_1$
$=1.5-1$
$=\frac{1}{2}$
View full question & answer→Question 142 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_0\cos^2\text{x}\text{ dx}$
Answer$\int\limits^\frac{\pi}{2}_0\cos^2\text{x}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{1+\cos2\text{x}}{2}\text{dx}$
$= \frac{1}{2}\int\limits^\frac{\pi}{2}_0(1+\cos2\text{x})\text{dx}$
$= \frac{1}{2}\Big[\text{x}+\frac{\sin2\text{x}}{2}\Big]^\frac{\pi}{2}_0$
$=\frac{1}{2}\Big[\frac{\pi}{2}+0\Big]$
$=\frac{\pi}{4}$
View full question & answer→Question 152 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}(\sin\text{x}+\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}(\sin\text{x}+\cos\text{x})\text{dx}$ Then,
$\text{I}=\big[-\cos\text{x}+\sin\text{x}\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0+1-(-1+0)$
$\Rightarrow\text{I}=2$
View full question & answer→Question 162 Marks
If f is an integrable function, show that:
$\int\limits^{\text{a}}_{-\text{a}}\text{xf}\big(\text{x}^2\big)\text{dx}=0$
Answer$\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{xf}(\text{x}^2)\text{dx}$
Let $\text{g(x)}=\text{f}\big(\text{x}^2\big)$
$\Rightarrow\text{g}(-\text{x})=(-\text{x})\text{f}(-\text{x})^2=-(\text{x})\text{f}(\text{x})^2=-\text{g(x)}\text{ i.e., g(x) is even}$
Therefore,
$\text{I}=0$ $\bigg[\text{Using}\int\limits^{\text{a}}_{-\text{a}}\text{g}(\text{x})\text{dx}=0\text{ when g(x) is odd}\bigg]$
View full question & answer→Question 172 Marks
Evaluate the following integrals:
$\int\limits^2_02\text{x}\big[\text{x}\big]\text{dx}$
Answer[x] = 0 for 0
and [x] = 1 for 1
Hence,
$=\int\limits^1_10+\int\limits^2_12\text{x}\text{ dx}$
$=\big\{\text{x}^2\big\}^2_1$
$=3$
View full question & answer→Question 182 Marks
Evaluate the following integrals:
$\int\limits^2_0\sqrt{4-\text{x}^2}\text{ dx}$
Answer$\int\limits^2_0\sqrt{4-\text{x}^2}\text{ dx}$
$=\int\limits^2_0\sqrt{2^2-\text{x}^2}\text{ dx}$
$=\Big[\frac{\text{x}}{2}\sqrt{4-\text{x}^2}\text{ dx}+\frac{1}{2}\times2^2\sin^{-1}\frac{\text{x}}{2}\Big]^2_0$
$=\Big[\frac{\text{x}}{2}\sqrt{4-\text{x}^2}\text{ dx}\Big]^2_0+2\Big[\sin^{-1}\frac{\text{x}}{2}\Big]^2_0$
$=0+2\Big(\frac{\pi}{2}-0\Big)$
$=\pi$
View full question & answer→Question 192 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\sin^{2}\text{x dx}$
Answer$\int\limits^\frac{\pi}{2}_0\sin^2\text{x}\text{dx}$
$= \int\limits^\frac{\pi}{2}_0\frac{1-\cos2\text{x}}{2}\text{dx}$
$= \frac{1}{2}\int\limits^\frac{\pi}{2}_0(1-\cos2\text{x})\text{dx}$
$=\frac{1}{2}\Big[\text{x}-\frac{\sin2\text{x}}{2}\Big]^\frac{\pi}{2}_0$
$=\frac{1}{2}(\frac{\pi}{2}-0)$
$=\frac{\pi}{4}$
View full question & answer→Question 202 Marks
Write the coefficient a, b, c of which the value of the integral $\int\limits^3_{-3}(\text{ax}^2+\text{bx}+\text{c})\text{dx}$ is independent.
Answer$\int\limits^3_{-3}(\text{ax}^2+\text{bx}+\text{c})\text{dx}$
$=\Big[\text{a}\frac{\text{x}^3}{3}+\text{b}\frac{\text{x}^2}{2}+\text{cx}\Big]^3_{-3}$
$=9\text{a}+\frac{9}{2}\text{b}+3\text{c}+9\text{a}-\frac{9}{2}\text{b}+3\text{c}$
$=18\text{a}+6\text{c}$
Hence, the given integral is independent of b.
View full question & answer→Question 212 Marks
If $\int_{0}^\limits{\text{a}}3\text{x}^2\text{ dx}=8,$ find the value of a.
AnswerWe have,
$\int_{0}^\limits{\text{a}}3\text{x}^2\text{ dx}=8$
$\Rightarrow\Big[3-\frac{\text{x}^3}{3}\Big]^{\text{a}}_0=8$
$\Rightarrow\text{a}^3=8$
$\Rightarrow\text{a}=2$
View full question & answer→Question 222 Marks
Evaluate the following integrals:
$\int\limits^{1}_0\text{e}^{\{\text{x}\}}\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{1}_0\text{e}^{\{\text{x}\}}\text{dx}$
We know that,
$\{\text{x}\}=\text{x},$ When $0<\text{x}<1$
$\therefore\ \text{I}=\int\limits^{1}_0\text{e}^{\{\text{x}\}}\text{dx}$
$=\big[\text{e}^{\text{x}}\big]^1_0$
$=\text{e}^1-\text{e}^0$
$=\text{e}-1$
View full question & answer→Question 232 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^3\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^3\text{x dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}(1-\cos^2\text{x})\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}\text{ dx}-\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}\cos^2\text{x dx}$
$=\big[-\cos\text{x}\big]^{\frac{\pi}{2}}_{\frac{-\pi}{2}}+\Big[\frac{\cos^3\text{x}}{3}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$=0+0=0$
View full question & answer→Question 242 Marks
Evaluate the following integrals:
$\int\limits^{\text{e}^2}_\text{e}\frac{1}{\text{x}\log\text{x}}\text{ dx}$
Answer$\int\limits^{\text{e}^2}_\text{e}\frac{1}{\text{x}\log\text{x}}\text{ dx}$
$=\int\limits^{\text{e}^2}_\text{e}\frac{\frac{1}{\text{x}}}{\log\text{x}}\text{ dx}$
$=\log\big[(\log{\text{x}})\big]^{\text{e}^2}_\text{e}$ $\Big[\int\frac{\text{f}'(\text{x})}{\text{f(x)}}\text{ dx}=\log\text{f(x)}+\text{C}\Big]$
$=\log\big(\log\text{e}^2\big)-\log(\log\text{e})$
$=\log(2\log\text{e})-\log(\log\text{e})$
$=\log2-\log1$ $(\log\text{e}=1)$
$=\log2-0$
$=\log2$
View full question & answer→Question 252 Marks
Evaluate the following integrals:
$\int\limits^2_0\big[\text{x}\big]\text{dx}$
Answerwe have,
$\text{I}=\int\limits^2_0\big[\text{x}\big]\text{dx}$
$=\int\limits^1_0\big[\text{x}\big]\text{dx}+\int\limits^2_1\big[\text{x}\big]\text{dx}$
$=\int\limits^1_00\text{ dx}+\int\limits^2_1(\text{1})\text{dx}$ $\begin{bmatrix}\because\big[\text{x}\big]=\begin{cases}0,&0\leq\text{x}<1\\1,&1\leq\text{x}<2\end{cases}\end{bmatrix}$
$=0+\big[\text{x}\big]^{2}_1$
$=2-1$
$=1$
View full question & answer→Question 262 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\text{x}\cos\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\text{x}\cos\text{x}\text{ dx}$ Then,
Integrating by parts
$\text{I}=\big[\text{x}\sin\text{x}\big]^{\frac{\pi}{2}}_0-\int_{0}^\limits{\frac{\pi}{2}}1\sin\text{x}\text{ dx}$
$\Rightarrow\text{I}=\big[\text{x}\sin\text{x}\big]^{\frac{\pi}{2}}_0+\big[\cos\text{x}\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=\frac{\pi}{2}-1$
View full question & answer→Question 272 Marks
Evaluate the following integrals:
$\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
Answer$\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
$=\int\limits^3_0\frac{1}{\text{x}^2+3^2}\text{ dx}$
$=\frac{1}{3}\Big[\tan^{-1}\frac{\text{x}}{3}\Big]^3_0$
$=\frac{1}{3}\big(\tan^{-1}1-\tan^{-1}0\big)$
$=\frac{1}{3}\Big(\frac{\pi}{4}-0\Big)$
$=\frac{\pi}{12}$
View full question & answer→Question 282 Marks
Evaluate the following integrals:
$\int\limits^1_{-2}\frac{|\text{x}|}{\text{x}}\text{ dx}$
AnswerLet $\int\limits^1_{-2}\frac{|\text{x}|}{\text{x}}\text{ dx}$
We have,
$|\text{x}|=\begin{cases}\text{x},&0\leq\text{x}\leq1\\-\text{x},&-2\leq\text{x}<0\end{cases}$
$\therefore\ \frac{|\text{x}|}{\text{x}}=\begin{cases}1,&0\leq\text{x}\leq1\\-1,&-2\leq\text{x}<0\end{cases}$
Therefore,
$\text{I}=\int\limits^0_{-2}-1\text{ dx}+\int\limits^1_01\text{ dx}$
$=-\big[\text{x}\big]^0_{-2}+\big[\text{x}\big]_0^1$
$=0-2+1-0$
$=-1$
View full question & answer→Question 292 Marks
Evaluate the following integrals:
$\int\limits^{4}_2\frac{\text{x}}{\text{x}^2+1}\text{dx}$
Answer$\int\limits^{4}_2\frac{\text{x}}{\text{x}^2+1}\text{dx}$
$=\frac{1}{2}\int\limits^{4}_2\frac{2\text{x}}{\text{x}^2+1}\text{dx}$
$=\frac{1}{2}\times\Big[\log(\text{x})^2-1\Big]^4_2$ $\Big[\int\frac{\text{f}'(\text{x})}{\text{f(x)}}\text{ dx}=\log\text{f(x)}+\text{C}\Big]$
$=\frac{1}{2}\big(\log17-\log5\big)$
$=\frac{1}{2}\log\Big(\frac{17}{5}\Big)$ $\Big(\log\text{a}-\log\text{b}=\log\frac{\text{a}}{\text{b}}\Big)$
View full question & answer→Question 302 Marks
Evaluate the following integrals:
$\int\limits^{1}_02^{\text{x}-[\text{x}]}\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{1}_02^{\text{x}-[\text{x}]}\text{dx}$
$=\int\limits^{1}_02^{\text{x}-0}\text{ dx}$ $\big(\because[\text{x}]=0,\text{ where}, 0<\text{x}<1\big)$
$=\int\limits^{1}_02^{\text{x}}\text{ dx}$
$=\Big[\frac{2^{\text{x}}}{\log_\text{e}2}\Big]^1_0$
$=\frac{2^1}{\log_\text{e}2}-\frac{2^0}{\log_\text{e}2}$
$=\frac{2}{\log_\text{e}2}-\frac{1}{\log_\text{e}2}$
$=\frac{1}{\log_\text{e}2}$
View full question & answer→Question 312 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}} \cos^2\text{x}\text{ dx}$
Answer$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}} \cos^2\text{x}\text{ dx}$
$= \int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\frac{1+\cos2\text{x}}{2}\text{dx}$
$=\frac{1}{2}\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}(1+\cos2\text{x})\text{dx}$
$=\frac{1}{2}\Big[\text{x}+\frac{\sin2\text{x}}{2}\Big]^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$=\frac{1}{2}\Big(\frac{\pi}{2}+0+\frac{\pi}{2}-0\Big)$
$=\frac{\pi}{2}$
View full question & answer→Question 322 Marks
If f is an integrable function, show that:
$\int\limits^{\text{a}}_{-\text{a}}\text{f}\big(\text{x}^2\big)\text{dx}=2\int\limits^\text{a}_0\text{f}\big(\text{x}^2\big)\text{dx}$
Answer$\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{f}\big(\text{x}^2\big)\text{dx}$
Here, $\text{g(x)}=\text{f}\big(\text{x}^2\big)$
$\Rightarrow\text{g}(-\text{x})=\text{f}(-\text{x})^2=\text{f}(\text{x})^2=\text{g(x)}\text{ i.e., g(x) is even}$
Therefore,
$\text{I}=2\int\limits^{\text{a}}_{0}\text{f}(\text{x}^2)\text{dx}$ $\bigg[\text{Using}\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{g}(\text{x})\text{dx}=\int\limits^{\text{a}}_{0}\text{f}(\text{x})\text{dx}\text{ when g(x) is even}\bigg]$
View full question & answer→Question 332 Marks
Evaluate the following integrals:
$\int\limits^{-\frac{\pi}{2}}_{-\frac{3\pi}{2}}\Big\{\sin^2(3\pi+\text{x})+(\pi+\text{x})^3\Big\}\text{dx}$
AnswerLet $\text{I}=\int\limits^{-\frac{\pi}{2}}_{-\frac{3\pi}{2}}\Big\{\sin^2(3\pi+\text{x})+(\pi+\text{x})^3\Big\}\text{dx}$
Put $\pi+\text{x}=\text{z}$
$\Rightarrow\text{dx}=\text{dz}$
When $\text{x}\rightarrow-\frac{3\pi}{2},\text{ z}\rightarrow-\frac{\pi}{2}$
When $\text{x}\rightarrow-\frac{\pi}{2},\text{ z}\rightarrow\frac{\pi}{2}$
$\therefore\ \text{I}=\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\Big[\sin^2(2\pi+\text{z})+\text{z}^3\Big]\text{dx}$
$=\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(\sin^2\text{z}+\text{z}^3\big)\text{dz}$ $\big[\sin(2\pi+\theta)=\sin\theta\big]$
$=\frac{1}{2}\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{1-\cos2\text{z}}{2}\text{ dz}+\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{z}^3\text{ dz}$
$=\frac{1}{2}\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dz}-\frac{1}{2}\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{z dz}+\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{z}^3\text{ dz}$
$=\frac{1}{2}\Big[\text{z}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}-\frac{1}{2}\Big[\frac{\sin2\text{z}}{2}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}+\Big[\frac{\text{z}^4}{4}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$=\frac{1}{2}\bigg[\frac{\pi}{2}-\Big(-\frac{\pi}{2}\Big)\bigg]-\frac{1}{4}\big[\sin\pi-\sin(-\pi)\big]+\frac{1}{4}\Big(\frac{\pi^4}{16}-\frac{\pi^4}{16}\big)$
$=\frac{1}{2}\times\pi-\frac{1}{4}(0+0)+\frac{1}{4}\times0$
$=\frac{\pi}{2}$
View full question & answer→Question 342 Marks
Evaluate the following definite integrals:
$\int_{-1}^\limits{1}\frac{1}{1+\text{x}^2} \text{ dx}$
AnswerLet $\text{I}=\int_{-1}^\limits{1}\frac{1}{1+\text{x}^2} \text{ dx}$ Then,
$\text{I}=\big[\tan^{-1}\text{x}\big]^1_{-1}$
$\Rightarrow\text{I}=\tan^{-1}1-\tan^{-1}(-1)$
$\Rightarrow\text{I}=\frac{\pi}{4}-\Big(-\frac{\pi}{4}\Big)$
$\Rightarrow\text{I}=\frac{\pi}{2}$
View full question & answer→Question 352 Marks
Evaluate the following definite integrals:
$\int_{4}\limits^{9}\frac{1}{\sqrt{\text{x}}} \text{ dx}$
AnswerLet $\text{I}=\int_{4}\limits^{9}\frac{1}{\sqrt{\text{x}}} \text{ dx}$ Then,
$\text{I}=2\int_{4}\limits^{9}\frac{1}{2\sqrt{\text{x}}} \text{ dx}$
$\Rightarrow\text{I}=2\big[\sqrt{\text{x}}\big]^9_4$
$\Rightarrow\text{I}=2(3-2)$
$\Rightarrow\text{I}=2$
View full question & answer→Question 362 Marks
If $\text{f(x)}=\int\limits^{\text{x}}_0\text{t}\sin\text{t dt},$ the write the value of f'(x).
Answer$\text{f(x)}=\int\limits^{\text{x}}_0\text{t}\sin\text{t dt}$
$\Rightarrow\text{f(x)}=\text{t}\big[-\cos\text{t}\big]^{\text{x}}_0-\int\limits^{\text{x}}_0\frac{\text{d}}{\text{dt}}(\text{t})\times(-\cos\text{t})\text{dt}$
$\Rightarrow\text{f(x)}=-(\text{x}\cos\text{x}-0)+\int\limits^{\text{x}}_0\cos\text{t dt}$
$\Rightarrow\text{f(x)}=-\text{x}\cos\text{x}+\big[\sin\text{t}\big]^{\text{x}}_0$
$\Rightarrow\text{f(x)}=-\text{x}\cos\text{x}+(\sin\text{x}-0)$
$\Rightarrow\text{f(x)}=-\text{x}\cos\text{x}+\sin\text{x}$
Differentiating both sides with respect to x, we get
$\text{f}'(\text{x})=-\big[\text{x}\times(-\sin\text{x})+\cos\text{x}\times1\big]+\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=-(-\text{x}\sin\text{x})-\cos\text{x}+\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=\text{x}\sin\text{x}$
Thus, the value of f'(x) is x $\sin\text{x}$
View full question & answer→Question 372 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\frac{\text{x}}{\text{x}+1}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{\text{x}}{\text{x}+1}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{1}1-\frac{1}{\text{x}+1}\text{ dx}$
$\Rightarrow\text{I}=\big[\text{x}-\log(\text{x}+1)\big]_0^1$
$\Rightarrow\text{I}=1-\log2-(0-\log1)$
$\Rightarrow\text{I}=\log\text{e}-\log2$
$\Rightarrow\text{I}=\log\frac{\text{e}}{2}$
View full question & answer→Question 382 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos\text{x}\text{ dx}$
AnswerWe have,
$\int\text{x}^2\cos\text{x dx}=\text{x}^2\int\cos\text{x dx}-\int(2\text{x})\big(\int\cos\text{x dx}\big)\text{dx}$
$=\text{x}^2\sin\text{x}-\int\sin\text{x }2\text{x dx}$
$=\text{x}^2\sin\text{x}-2\big[-\text{x}\cos\text{x}+\int\cos\text{x dx}\big]$
$\therefore\ \int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos\text{x}\text{ dx}=\big[\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x}\big]^{\frac{\pi}{2}}_0$
$=\Big[\frac{\pi}{4}+0-2-0-0+0\Big]$
$=\frac{\pi^2}{4}-2$
View full question & answer→Question 392 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\frac{1}{1+\text{x}^2} \text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{1}{1+\text{x}^2} \text{ dx}$ Then,
$\text{I}=\big[\tan^{-1}\text{x}\big]^1_0$
$\Rightarrow\text{I}=\tan^{-1}1-\tan^{-1}0$
$\Rightarrow\text{I}=\frac{\pi}{4}-0$
$\Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 402 Marks
Evaluate the following integrals:
$\int\limits^\text{x}_{0}\text{e}^{-\text{x}}\text{ dx}$
Answer$\int\limits^\text{x}_{0}\text{e}^{-\text{x}}\text{ dx}$
$=-\big[\text{e}^{-\text{x}}\big]^{\text{x}}_0$
$=-(0-1)$
$=0+1$
$=1$
View full question & answer→Question 412 Marks
Evaluate the following integrals:
$\int\limits^3_2\frac{1}{\text{x}}\text{ dx}$
Answer$\int\limits^3_2\frac{1}{\text{x}}\text{ dx}$
$=\Big[\log_\text{e}\text{x}\Big]^3_2$
$=\log_\text{e}3-\log_\text{e}2$
$=\log_\text{e}\Big(\frac{3}{2}\Big)$
View full question & answer→Question 422 Marks
Evaluate the following integrals:
$\int\limits^4_0\frac{1}{\sqrt{16-\text{x}^2}}\text{ dx}$
Answer$\int\limits^4_0\frac{1}{\sqrt{16-\text{x}^2}}\text{ dx}$
$=\int\limits^4_0\frac{1}{\sqrt{14^2-\text{x}^2}}\text{ dx}$
$=\Big[\sin^{-1}\frac{\text{x}}{4}\Big]^4_0$
$=\Big(\frac{\pi}{2}-0\Big)$
$=\frac{\pi}{2}$
View full question & answer→Question 432 Marks
Evaluate the following definite integrals:
$\int_{-2}^\limits{3}\frac{1}{\text{x}+7} \text{ dx}$
AnswerWe know that $\int\frac{\text{dx}}{\text{x}}=\log\text{x}+\text{C}$
Now,
$\int_{-2}^\limits{3}\frac{1}{\text{x}+7} \text{ dx}$
$=\big[\log(\text{x}+7)\big]^3_{-2}$
$=\big[\log10-\log5]^3_{-2}$
$=\log\frac{10}{5}$ $\Big[\because\log\text{a}-\log\text{b}=\log\frac{\text{a}}{\text{b}}\Big]$
$=\log2$
$\therefore\ \int_{-2}^\limits{3}\frac{1}{\text{x}+7} \text{ dx}=\log2$
View full question & answer→Question 442 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{4}_0\sin2\text{x dx}$
Answer$\int\limits^\frac{\pi}{4}_0\sin2\text{x dx}$
$=\Big[\frac{-\cos2\text{x}}{2}\Big]^{\frac{\pi}{4}}_0$
$=-\frac{1}{2}\Big(\cos\frac{\pi}{2}-\cos0\Big)$
$=-\frac{1}{2}\times(0-1)$
$=\frac{1}{2}$
View full question & answer→Question 452 Marks
Evaluate the following integrals:
$\int\limits^1_0\frac{1}{1+\text{x}^2}\text{ dx}$
Answer$\int\limits^1_0\frac{1}{1+\text{x}^2}\text{ dx}$
$=\big[\tan^{-1}\text{x}\big]^1_0$
$=\tan^{-1}1-\tan^{-1}0$
$=\frac{\pi}{4}-0$
$=\frac{\pi}{4}$
View full question & answer→Question 462 Marks
If $\int\limits^{\text{a}}_03\text{x}^2\text{ dx}=8,$ Write the value of a.
AnswerWe have,
$\int\limits^{\text{a}}_03\text{x}^2\text{ dx}=8$
$\Rightarrow\Big[3\frac{\text{x}^3}{3}\Big]^{\text{a}}_0=8$
$\Rightarrow\big[\text{x}^3\big]^{\text{a}}_0=8$
$\Rightarrow\text{a}^3-0=8$
$\Rightarrow\text{a}=\sqrt[3]{8}$
$\Rightarrow\text{a}=2$
View full question & answer→Question 472 Marks
Evaluate the following definite integrals:
$\int_{-2}\limits^{\frac{1}{2}}\frac{1}{\sqrt{1-\text{x}^2}} \text{ dx}$
AnswerLet $\int_{-2}\limits^{\frac{1}{2}}\frac{1}{\sqrt{1-\text{x}^2}} \text{ dx}$ Then,
$\text{I}=\big[\sin^{-1}\text{x}\big]^{\frac{1}{2}}_0$
$\Rightarrow\text{I}=\sin^{-1}\frac{1}{2}-\sin^{-1}0$
$\Rightarrow\text{I}=\frac{\pi}{6}-0$
$\Rightarrow\text{I}=\frac{\pi}{6}$
View full question & answer→Question 482 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$
AnswerLet $\text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$
Here, $\text{f}(\theta)=\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)$
Consider, $\text{f}(-\theta)=\log\bigg[\frac{\text{a}-\sin(-\theta)}{\text{a}+\sin(-\theta)}\bigg]$
$=-\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)=-\text{f}(\theta)$
i.e., $\text{f}(\theta)$ is odd function.
Therefore, $\text{I}=0$
View full question & answer→Question 492 Marks
Evaluate the following integrals:
$\int\limits^1_0\text{xe}^{\text{x}^2}\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^1_0\text{xe}^{\text{x}^2}\text{dx}=\frac{1}{2}\int\limits^1_0\text{e}^{\text{x}^2}2\text{dx}$
Putting $\text{x}^2=\text{z}$
$2\text{x dx}=\text{dz}$
When $\text{x}\rightarrow0;\text{ z}\rightarrow0$
And $\text{x}\rightarrow1;\text{ z}\rightarrow1$
$\therefore\ \text{I}=\frac{1}{2}\int\limits^1_0\text{e}^2\text{ dz}$
$=\frac{1}{2}\times\big[\text{e}^{\text{z}}\big]^1_0$
$=\frac{1}{2}(\text{e}-\text{e}^0)$
$=\frac{1}{2}(\text{e}-1)$
View full question & answer→Question 502 Marks
Evaluate the following integrals:
$\int\limits^{2}_0\text{x}[\text{x}]\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{2}_0\text{x}[\text{x}]\text{dx}$
We know that,
$\text{x}[\text{x}]=\begin{cases}\text{x}\times0,&0<\text{x}<1\\\text{x}\times1,&1<\text{x}<2\end{cases}$
i.e.,
$\text{x}[\text{x}]=\begin{cases}0,&0<\text{x}<1\\\text{x},&1<\text{x}<2\end{cases}$
$\therefore\ \text{I}=\int\limits^{2}_0\text{x}[\text{x}]\text{dx}$
$=\int\limits^{1}_0\text{x}[\text{x}]\text{dx}+\int\limits^{2}_1\text{x}[\text{x}]\text{dx}$
$=\int\limits^{1}_0{0}\text{ dx}+\int\limits^{2}_1(\text{x})\text{dx}$
$=0+\Big[\frac{\text{x}^2}{2}\Big]^2_1$
$=\frac{2^2}{2}-\frac{1^2}{2}$
$=\frac{3}{2}$
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