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

    $\displaystyle\Big[ {{x^5}\over{5}}-{{2\,x^3}\over{3}}+x \Big]_0^1 = {{8}\over{15}}$ $\displaystyle\Big[ {{x^3}\over{3}} \Big]_0^1 = {{1}\over{3}}$ $\displaystyle\Big[ x-{{x^3}\over{3}} \Big]_0^1 = {{2}\over{3}}$ $\displaystyle\Big[ x^4-2\,x^2+1 \Big]_0^1 = -1$

  2. Find $\displaystyle \int {{{1-\cos ^2x}\over{\cos ^2x}}}{\;dx}$.

    $\displaystyle -\tan x + C$ $\displaystyle \tan x-x + C$ $\displaystyle \tan x+x + C$ $\displaystyle \tan x + C$

  3. Find $\displaystyle \int {{{\left(\ln x\right)^2}\over{x}}}{\;dx}$ by substituting $\displaystyle u=\ln x$

    $\displaystyle \int {u^2}{\;du} = {{\left(\ln x\right)^3}\over{3}} + C$ $\displaystyle \int {u}{\;du} = {{\left(\ln x\right)^2}\over{2}} + C$ $\displaystyle 2\,\int {u}{\;du} = \left(\ln x\right)^2 + C$ $\displaystyle \int {e^{u}}{\;du} = x + C$

  4. Find $\displaystyle \int {\left(\sec x\right)^3\,\tan x}{\;dx}$ by using substitution.

    $\displaystyle -{{1}\over{2\,\cos ^2x}} + C$ $\displaystyle -{{1}\over{\cos x}} + C$ $\displaystyle {{1}\over{3\,\cos ^3x}} + C$ $\displaystyle {{1}\over{2\,\cos ^2x}} + C$

  5. Evaluate $\displaystyle \int_{1}^{4}{{{1}\over{\sqrt{x}}}\;dx}$.

    $\displaystyle\Big[ -{{1}\over{2\,x^{{{3}\over{2}}}}} \Big]_1^4 = {{7}\over{16}}$ $\displaystyle\Big[ {{1}\over{2\,\sqrt{x}}} \Big]_1^4 = -{{1}\over{4}}$ $\displaystyle\Big[ {{1}\over{\sqrt{x}}} \Big]_1^4 = -{{1}\over{2}}$ $\displaystyle\Big[ 2\,\sqrt{x} \Big]_1^4 = 2$ $\displaystyle\Big[ \sqrt{x} \Big]_1^4 = 1$

  6. Find $\displaystyle \int {{{\left(x+1\right)^2}\over{\sqrt{x}}}}{\;dx}$.

    $\displaystyle {{2\,x^{{{5}\over{2}}}}\over{5}}+{{4\,x^{{{3}\over{2}}}}\over{3}}+2 \,\sqrt{x} + C$ $\displaystyle {{2\,x^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{2\,\left(x+1\right)}\over{\sqrt{x}}}-{{\left(x+1\right)^2}\over{2 \,x^{{{3}\over{2}}}}} + C$ $\displaystyle {{2\,x^{{{7}\over{2}}}}\over{7}} + C$

  7. Find $\displaystyle \int {{{x^2}\over{\sqrt{x^3+1}}}}{\;dx}$ by substituting $\displaystyle u=x^3+1$

    $\displaystyle \int {{{1}\over{\sqrt{u}}}}{\;du} = 2\,\sqrt{x^3+1} + C$ $\displaystyle \int {{{1}\over{u^{{{3}\over{2}}}}}}{\;du} = -{{2}\over{\sqrt{x^3+1}}} + C$ $\displaystyle {{\int {{{1}\over{\sqrt{u}}}}{\;du}}\over{3}} = {{2\,\sqrt{x^3+1}}\over{3}} + C$ $\displaystyle {{\int {{{u-1}\over{\sqrt{u}}}}{\;du}}\over{3}} = {{2\,\left(x^3+1\right)^{{{3}\over{2}}}}\over{9}}-{{2\,\sqrt{x^3+1} }\over{3}} + C$

  8. Evaluate $\displaystyle \int_{-2}^{-1}{e^{x+2}\;dx}$.

    $\displaystyle\Big[ e^{x+3} \Big]_{ -2}^{ -1} = e^2-e$ $\displaystyle\Big[ e^{x+2} \Big]_{ -2}^{ -1} = e-1$ $\displaystyle\Big[ e^{x} \Big]_{ -2}^{ -1} = e^ {- 1 }-e^ {- 2 }$ $\displaystyle\Big[ 2\,e^{x} \Big]_{ -2}^{ -1} = 2\,e^ {- 1 }-2\,e^ {- 2 }$

  9. Find $\displaystyle \int {\left(x-2\right)^{{{3}\over{2}}}}{\;dx}$ by using substitution.

    $\displaystyle -{{2\,\left(x-2\right)^{{{5}\over{2}}}}\over{5}} + C$ $\displaystyle {{2\,\left(x-2\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{2\,\left(x-2\right)^{{{5}\over{2}}}}\over{5}} + C$ $\displaystyle {{2\,\left(x-2\right)^{{{5}\over{2}}}}\over{5}} + C$ $\displaystyle -{{2\,\left(x-2\right)^{{{5}\over{2}}}}\over{5}} + C$

  10. Find $\displaystyle \int {{{e^{\sqrt{x}}}\over{\sqrt{x}}}}{\;dx}$ by using substitution.

    $\displaystyle 2\,e^{\sqrt{x}} + C$ $\displaystyle x + C$ $\displaystyle -2\,e^{\sqrt{x}} + C$ $\displaystyle -x + C$


Department of Mathematics
Last modified: 2024-09-08