Find $\displaystyle\displaystyle \displaystyle\int{{{x}^{2}}}\sqrt{{{x}^{3}}+1}\ \mathrm{d}x$.[2]

Question

Find $\displaystyle\displaystyle \displaystyle\int{{{x}^{2}}}\sqrt{{{x}^{3}}+1}\ \mathrm{d}x$.[2] Hence, find the exact value of $\displaystyle\displaystyle \displaystyle\int_{-1}^{0}{{{x}^{5}}}\sqrt{{{x}^{3}}+1}\ \mathrm{d}x$.[3] Use the substitution $u=1+{{\mathrm{e}}^{x}}$ to find $\displaystyle\displaystyle \displaystyle\int{{{\mathrm{e}}^{2x}}}\sqrt{1+{{\mathrm{e}}^{x}}}\ \mathrm{d}x$.[4]

Accepted Answer

(a)(i) $\frac29(x^3+1)^{3/2}+C$ (ii) $-\frac4{45}$ (b) $\frac25(1+\mathrm e^x)^{5/2}-\frac23(1+\mathrm e^x)^{3/2}+C$

2023 YIJC P1 Q9

2023 YIJC P1 Q9

Video Solution:

Solution:

Solution locked