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

    $\displaystyle \int_{0}^{1}{{{y^2-6\,y+9}\over{2}}-{{y^2}\over{2}}\;dy}$ = 3 $\displaystyle \int_{0}^{1}{y^2-3\,y+3\;dy}$ = $\displaystyle {{11}\over{6}}$ $\displaystyle \int_{0}^{1}{\left(3-y\right)^3\;dy}$ = $\displaystyle {{65}\over{4}}$ $\displaystyle \int_{0}^{1}{\left(3-y\right)^2\;dy}$ = $\displaystyle {{19}\over{3}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

    $\displaystyle \int_{-1}^{1}{{{x^4+2\,x^3+2\,x^2+2\,x+1}\over{2}}-2\,x^4-2\,x^3\;d x}$ = $\displaystyle {{16}\over{15}}$ $\displaystyle \int_{-1}^{1}{3\,x^4-x^3-2\,x^2+x-1\;dx}$ = $\displaystyle -{{32}\over{15}}$ $\displaystyle \int_{-3}^{3}{3\,x^4-x^3-2\,x^2+x-1\;dx}$ = $\displaystyle {{1248}\over{5}}$ $\displaystyle \int_{-3}^{3}{{{x^4+2\,x^3+2\,x^2+2\,x+1}\over{2}}-2\,x^4-2\,x^3\;d x}$ = $\displaystyle -{{624}\over{5}}$

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

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

  10. Evaluate $\displaystyle \int_{0}^{\pi}{\int_{0}^{{{\pi}\over{2}}}{2\,\cos x-2\,\sin y\;dy} \;dx}$.

    $\displaystyle \int_{0}^{\pi}{\pi\,\cos x-2\;dx}$ = $\displaystyle -2\,\pi$ $\displaystyle \int_{0}^{\pi}{\pi\,\sin x+2\;dx}$ = $\displaystyle -2\,\pi$ $\displaystyle \int_{0}^{\pi}{2-\pi\,\sin x\;dx}$ = $\displaystyle -2\,\pi$ $\displaystyle \int_{0}^{\pi}{-\pi\,\cos x-2\;dx}$ = $\displaystyle -2\,\pi$


Department of Mathematics
Last modified: 2026-09-24