Evaluate $\displaystyle \int_{0}^{1}{\int_{1}^{4}{{{y-x}\over{\sqrt{y}}}\;dy}\;dx}$ .
$\displaystyle \int_{0}^{1}{{{\left(4-x\right)^2}\over{4}}-\left(1-x\right)^2\;dx}$ = $\displaystyle {{11}\over{3}}$ $\displaystyle \int_{0}^{1}{{{2\,x-1}\over{2}}-4\,x+8\;dx}$ = $\displaystyle {{11}\over{3}}$ $\displaystyle -{{\int_{0}^{1}{6\,x-14\;dx}}\over{3}}$ = $\displaystyle {{11}\over{3}}$ $\displaystyle -2\,\int_{0}^{1}{x\;dx}$ = $\displaystyle {{11}\over{3}}$
Evaluate $\displaystyle \int_{0}^{1}{\int_{0}^{1}{\left(2\,x-y\right)^2\;dx}\;dy}$ .
$\displaystyle \int_{0}^{1}{{{y^2-4\,y+4}\over{2}}-{{y^2}\over{2}}\;dy}$ = 1 $\displaystyle \int_{0}^{1}{\left(2-y\right)^2\;dy}$ = $\displaystyle {{7}\over{3}}$ $\displaystyle {{\int_{0}^{1}{3\,y^2-6\,y+4\;dy}}\over{3}}$ = $\displaystyle {{2}\over{3}}$ $\displaystyle \int_{0}^{1}{\left(2-y\right)^3\;dy}$ = $\displaystyle {{15}\over{4}}$
Evaluate $\displaystyle\int\!\!\int_D x\,y dxdy$ where is the region bounded by $\displaystyle x=y^2$ and x = 4y − 3.
$\displaystyle \int_{1}^{3}{y\,\left({{16\,y^2-24\,y+9}\over{2}}-{{y^4}\over{2}} \right)\;dy}$ = $\displaystyle {{40}\over{3}}$ $\displaystyle \int_{0}^{2}{y\,\left({{16\,y^2-24\,y+9}\over{2}}-{{y^4}\over{2}} \right)\;dy}$ = $\displaystyle {{11}\over{3}}$ $\displaystyle -\int_{0}^{2}{y\,\left({{16\,y^2-24\,y+9}\over{2}}-{{y^4}\over{2}} \right)\;dy}$ = $\displaystyle -{{11}\over{3}}$ $\displaystyle -\int_{1}^{3}{y\,\left({{16\,y^2-24\,y+9}\over{2}}-{{y^4}\over{2}} \right)\;dy}$ = $\displaystyle -{{40}\over{3}}$
Evaluate $\displaystyle \int_{1}^{2}{y\,\int_{1}^{2}{x\,\sqrt{y^2+x^2}\;dx}\;dy}$ .
$\displaystyle \int_{1}^{2}{y\,\left(2\,\sqrt{y^2+4}-2\,\sqrt{y^2+1}\right)\;dy}$ = $\displaystyle -{{2\,5^{{{3}\over{2}}}-2^{{{11}\over{2}}}}\over{3}}-{{2\,5^{{{3 }\over{2}}}-2^{{{5}\over{2}}}}\over{3}}$ $\displaystyle \int_{1}^{2}{y\,\left(y^2+4\right)^{{{3}\over{2}}}\;dy}$ = $\displaystyle {{2^{{{15}\over{2}}}}\over{5}}-5^{{{3}\over{2}}}$ $\displaystyle \int_{1}^{2}{y\,\sqrt{y^2+4}\;dy}$ = $\displaystyle {{2^{{{9}\over{2}}}}\over{3}}-{{5^{{{3}\over{2}}}}\over{3}}$ $\displaystyle \int_{1}^{2}{y\,\left({{\left(y^2+4\right)^{{{3}\over{2}}}}\over{3 }}-{{\left(y^2+1\right)^{{{3}\over{2}}}}\over{3}}\right)\;dy}$ = $\displaystyle -{{5^{{{5}\over{2}}}-2^{{{15}\over{2}}}}\over{15}}-{{5^{{{5}\over{2 }}}-2^{{{5}\over{2}}}}\over{15}}$
Evaluate $\displaystyle \int_{0}^{{{\pi}\over{4}}}{\int_{0}^{{{\pi}\over{4}}}{\sin \left(y+ x\right)\;dx}\;dy}$ .
$\displaystyle \int_{0}^{{{\pi}\over{4}}}{\sin y\;dy}$ = $\displaystyle 1-{{1}\over{\sqrt{2}}}$ $\displaystyle \int_{0}^{{{\pi}\over{4}}}{\cos y\;dy}$ = $\displaystyle {{1}\over{\sqrt{2}}}$ $\displaystyle \int_{0}^{{{\pi}\over{4}}}{\cos y-\cos \left({{4\,y+\pi}\over{4}} \right)\;dy}$ = $\displaystyle \sqrt{2}-1$ $\displaystyle \left(1-{{1}\over{\sqrt{2}}}\right)\,\int_{0}^{{{\pi}\over{4}}}{ \sin y\;dy}$ = $\displaystyle \left(1-{{1}\over{\sqrt{2}}}\right)^2$
Evaluate $\displaystyle \int_{1}^{2}{\int_{1}^{4}{{{y+x}\over{\sqrt{x}}}\;dx}\;dy}$ .
$\displaystyle \int_{1}^{2}{{{y+4}\over{2}}-y-1\;dy}$ = $\displaystyle {{5}\over{32}}$ $\displaystyle {{\int_{1}^{2}{\ln 4\,y+3\;dy}}\over{2}}$ = $\displaystyle {{23}\over{3}}$ $\displaystyle \int_{1}^{2}{{{1}\over{2}}-{{y+4}\over{16}}\;dy}$ = $\displaystyle {{5}\over{32}}$ $\displaystyle {{\int_{1}^{2}{6\,y+14\;dy}}\over{3}}$ = $\displaystyle {{23}\over{3}}$
Evaluate $\displaystyle\int\!\!\int_D 2\,y+x dxdy$ where is the region bounded by $\displaystyle y=2\,x^2$ and $\displaystyle y=x^2+9$ .
$\displaystyle \int_{-1}^{1}{-3\,x^4-x^3+18\,x^2+9\,x+81\;dx}$ = $\displaystyle {{864}\over{5}}$ $\displaystyle \int_{-3}^{3}{-{{x^4-2\,x^3+18\,x^2-18\,x+81}\over{2}}+2\,x^4-2\,x^ 3\;dx}$ = $\displaystyle -{{1296}\over{5}}$ $\displaystyle \int_{-1}^{1}{-{{x^4-2\,x^3+18\,x^2-18\,x+81}\over{2}}+2\,x^4-2\,x^ 3\;dx}$ = $\displaystyle -{{432}\over{5}}$ $\displaystyle \int_{-3}^{3}{-3\,x^4-x^3+18\,x^2+9\,x+81\;dx}$ = $\displaystyle {{2592}\over{5}}$
Evaluate $\displaystyle \int_{0}^{1}{\int_{0}^{1}{\left(2\,y-x^2\right)^2\;dy}\;dx}$ .
$\displaystyle \int_{0}^{1}{x^4\;dx}$ = $\displaystyle {{13}\over{15}}$ $\displaystyle \int_{0}^{1}{\left(2-x^2\right)^2-x^4\;dx}$ = $\displaystyle {{13}\over{15}}$ $\displaystyle {{\int_{0}^{1}{3\,x^4-6\,x^2+4\;dx}}\over{3}}$ = $\displaystyle {{13}\over{15}}$ $\displaystyle \int_{0}^{1}{1-x^2\;dx}$ = $\displaystyle {{13}\over{15}}$
Evaluate $\displaystyle \int_{0}^{1}{\int_{-3}^{-2}{e^{x+3}\,y^2+x\;dx}\;dy}$ .
$\displaystyle {{\int_{0}^{1}{\left(2\,e-2\right)\,y^2-5\;dy}}\over{2}}$ = $\displaystyle {{2\,e-17}\over{6}}$ $\displaystyle \int_{0}^{1}{3\,e^ {- 2 }\,y^2-3\,e^ {- 3 }\,y^2+1\;dy}$ = $\displaystyle {{2\,e-17}\over{6}}$ $\displaystyle \int_{0}^{1}{e^ {- 2 }\,y^2-e^ {- 3 }\,y^2+1\;dy}$ = $\displaystyle {{2\,e-17}\over{6}}$ $\displaystyle \int_{0}^{1}{e^2\,y^2-e\,y^2+1\;dy}$ = $\displaystyle {{2\,e-17}\over{6}}$
Evaluate $\displaystyle \int_{0}^{{{\pi}\over{2}}}{\int_{0}^{\pi}{-2\,\sin y-\cos x\;dy}\;d x}$ .
$\displaystyle -2\,\pi\,\int_{0}^{{{\pi}\over{2}}}{\sin x\;dx}$ = $\displaystyle -3\,\pi$ $\displaystyle \int_{0}^{{{\pi}\over{2}}}{\pi\,\cos x-4\;dx}$ = $\displaystyle -3\,\pi$ $\displaystyle 2\,\pi\,\int_{0}^{{{\pi}\over{2}}}{\sin x\;dx}$ = $\displaystyle -3\,\pi$ $\displaystyle \int_{0}^{{{\pi}\over{2}}}{-\pi\,\cos x-4\;dx}$ = $\displaystyle -3\,\pi$

Department of Mathematics
Last modified: 2025-07-29