1. Evaluate $\displaystyle \int_{0}^{1}{\int_{0}^{1}{\left(3\,y-x^2\right)^2\;dy}\;dx}$.

    $\displaystyle \int_{0}^{1}{x^4\;dx}$ = $\displaystyle {{11}\over{5}}$ $\displaystyle \int_{0}^{1}{x^4-3\,x^2+3\;dx}$ = $\displaystyle {{11}\over{5}}$ $\displaystyle -{{\int_{0}^{1}{2\,x^2-3\;dx}}\over{2}}$ = $\displaystyle {{11}\over{5}}$ $\displaystyle \int_{0}^{1}{\left(3-x^2\right)^2-x^4\;dx}$ = $\displaystyle {{11}\over{5}}$

  2. Evaluate $\displaystyle \int_{0}^{1}{\int_{0}^{1}{e^{y+x}\;dy}\;dx}$.

    $\displaystyle \left(e-1\right)\,\int_{0}^{1}{x\;dx}$ = $\displaystyle {{e-1}\over{2}}$ $\displaystyle \int_{0}^{1}{e^{x+1}-e^{x}\;dx}$ = $\displaystyle e^2-2\,e+1$ $\displaystyle {{\int_{0}^{1}{e^{x}\;dx}}\over{2}}$ = $\displaystyle {{e-1}\over{2}}$ $\displaystyle \int_{0}^{1}{x\,e^{x}\;dx}$ = 1

  3. Find the volume of the tetrahedron bounded by z = 2 − 2y, y = x, $x=0$ and $z=0$.

    $\displaystyle \int_{0}^{1}{x^2-4\,x+3\;dx}$ = $\displaystyle {{4}\over{3}}$ $\displaystyle \int_{0}^{2}{x^2-2\,x+1\;dx}$ = $\displaystyle {{2}\over{3}}$ $\displaystyle \int_{0}^{2}{x^2-4\,x+3\;dx}$ = $\displaystyle {{2}\over{3}}$ $\displaystyle \int_{0}^{1}{x^2-2\,x+1\;dx}$ = $\displaystyle {{1}\over{3}}$

  4. Evaluate $\displaystyle\int\!\!\int_D 2\,x\,y dxdy$ where $D$ is the region bounded by $\displaystyle x=y^2$ and x = 1.

    $\displaystyle \int_{-1}^{1}{y\,\left({{1}\over{2}}-{{y^4}\over{2}}\right)\;dy}$ = 0 $\displaystyle \int_{-2}^{0}{y\,\left({{1}\over{2}}-{{y^4}\over{2}}\right)\;dy}$ = $\displaystyle {{13}\over{3}}$ $\displaystyle 2\,\int_{-1}^{1}{y\,\left({{1}\over{2}}-{{y^4}\over{2}}\right)\;dy}$ = 0 $\displaystyle 2\,\int_{-2}^{0}{y\,\left({{1}\over{2}}-{{y^4}\over{2}}\right)\;dy}$ = $\displaystyle {{26}\over{3}}$

  5. Evaluate $\displaystyle \int_{0}^{1}{\int_{0}^{1}{\left(y+3\,x\right)^2\;dx}\;dy}$.

    $\displaystyle \int_{0}^{1}{y^2+3\,y+3\;dy}$ = $\displaystyle {{29}\over{6}}$ $\displaystyle \int_{0}^{1}{\left(y+3\right)^3\;dy}$ = $\displaystyle {{175}\over{4}}$ $\displaystyle \int_{0}^{1}{\left(y+3\right)^2\;dy}$ = $\displaystyle {{37}\over{3}}$ $\displaystyle \int_{0}^{1}{{{y^2+6\,y+9}\over{2}}-{{y^2}\over{2}}\;dy}$ = 6

  6. Evaluate $\displaystyle \int_{0}^{1}{\int_{1}^{4}{{{y-x}\over{\sqrt{y}}}\;dy}\;dx}$.

    $\displaystyle -2\,\int_{0}^{1}{x\;dx}$ = $\displaystyle {{11}\over{3}}$ $\displaystyle \int_{0}^{1}{{{\left(4-x\right)^2}\over{4}}-\left(1-x\right)^2\;dx}$ = $\displaystyle {{11}\over{3}}$ $\displaystyle \int_{0}^{1}{{{2\,x-1}\over{2}}-4\,x+8\;dx}$ = $\displaystyle {{11}\over{3}}$ $\displaystyle -{{\int_{0}^{1}{6\,x-14\;dx}}\over{3}}$ = $\displaystyle {{11}\over{3}}$

  7. Evaluate $\displaystyle \int_{1}^{2}{y\,\int_{1}^{2}{{{x}\over{\sqrt{y^2+x^2}}}\;dx}\;dy}$.

    $\displaystyle \int_{1}^{2}{y\,\left(\sqrt{y^2+4}-\sqrt{y^2+1}\right)\;dy}$ = $\displaystyle -{{5^{{{3}\over{2}}}-2^{{{9}\over{2}}}}\over{3}}-{{5^{{{3}\over{2}} }-2^{{{3}\over{2}}}}\over{3}}$ $\displaystyle \int_{1}^{2}{y\,\sqrt{y^2+4}\;dy}$ = $\displaystyle {{2^{{{9}\over{2}}}}\over{3}}-{{5^{{{3}\over{2}}}}\over{3}}$ $\displaystyle \int_{1}^{2}{y\,\left({{2}\over{\sqrt{y^2+1}}}-{{2}\over{\sqrt{y^2+ 4}}}\right)\;dy}$ = $\displaystyle 4\,\sqrt{5}-3\,2^{{{3}\over{2}}}$ $\displaystyle \int_{1}^{2}{{{y}\over{\sqrt{y^2+4}}}\;dy}$ = $\displaystyle 2^{{{3}\over{2}}}-\sqrt{5}$

  8. Evaluate $\displaystyle \int_{0}^{{{\pi}\over{4}}}{\int_{0}^{{{\pi}\over{4}}}{\sin \left(y+ x\right)\;dx}\;dy}$.

    $\displaystyle \int_{0}^{{{\pi}\over{4}}}{\cos y-\cos \left({{4\,y+\pi}\over{4}} \right)\;dy}$ = $\displaystyle \sqrt{2}-1$ $\displaystyle \left(1-{{1}\over{\sqrt{2}}}\right)\,\int_{0}^{{{\pi}\over{4}}}{ \sin y\;dy}$ = $\displaystyle \left(1-{{1}\over{\sqrt{2}}}\right)^2$ $\displaystyle \int_{0}^{{{\pi}\over{4}}}{\sin y\;dy}$ = $\displaystyle 1-{{1}\over{\sqrt{2}}}$ $\displaystyle \int_{0}^{{{\pi}\over{4}}}{\cos y\;dy}$ = $\displaystyle {{1}\over{\sqrt{2}}}$

  9. Evaluate $\displaystyle \int_{1}^{2}{\int_{1}^{4}{{{y-x}\over{\sqrt{x}}}\;dx}\;dy}$.

    $\displaystyle \int_{1}^{2}{-{{y-4}\over{16}}-{{1}\over{2}}\;dy}$ = $\displaystyle -{{11}\over{32}}$ $\displaystyle {{\int_{1}^{2}{6\,y-14\;dy}}\over{3}}$ = $\displaystyle -{{5}\over{3}}$ $\displaystyle \int_{1}^{2}{-y+{{y-4}\over{2}}+1\;dy}$ = $\displaystyle -{{11}\over{32}}$ $\displaystyle {{\int_{1}^{2}{\ln 4\,y-3\;dy}}\over{2}}$ = $\displaystyle -{{5}\over{3}}$

  10. Evaluate $\displaystyle\int\!\!\int_D 2\,y+x dxdy$ where $D$ is the region bounded by $\displaystyle y=2\,x^2$ and $\displaystyle y=x^2+4$.

    $\displaystyle \int_{-3}^{3}{-3\,x^4-x^3+8\,x^2+4\,x+16\;dx}$ = $\displaystyle -{{258}\over{5}}$ $\displaystyle \int_{-2}^{2}{-3\,x^4-x^3+8\,x^2+4\,x+16\;dx}$ = $\displaystyle {{1024}\over{15}}$ $\displaystyle \int_{-3}^{3}{3\,x^4-x^3-8\,x^2+4\,x-16\;dx}$ = $\displaystyle {{258}\over{5}}$ $\displaystyle \int_{-2}^{2}{3\,x^4-x^3-8\,x^2+4\,x-16\;dx}$ = $\displaystyle -{{1024}\over{15}}$


Department of Mathematics
Last modified: 2025-10-05