Question

Evaluate:
i. ${\int }_0^{\frac{\pi }{2}}\text{(sin x + cos x)dx}$
ii. ${\int }_0^1(3x^2+4)dx$
iii. ${\int }_0^1(3e^{3x}+e^{-x})dx$

• A. i. 0; ii. 5; iii. $\frac{e^4-1}{e}$
• B. i. 0; ii. 5; iii. $e^3-1$
• C. i. 2; ii. 5; iii. $e^3-1$
• D. i. 2; ii. 5; iii. $\frac{e^4-1}{e}$

Answer 1

The correct option is D i. 2; ii. 5; iii. $\frac{e^4-1}{e}$

i. ${\int }_0^{\frac{\pi }{2}}\text{(sin x + cos x)dx}$
$={[-cosx]}_0^{\frac{\pi }{2}}+{\left[sinx\right]}_0^{\frac{\pi }{2}}=(-0+1)+(1-0)=2$

ii. ${\int }_0^1(3x^2+4)dx=[x^3{]}_0^1+[4x{]}_0^1=1+4=5$

iii.${\int }_0^1(3e^{3x}+e^{-x})dx=[e^{3x}{]}_0^1+[-e^{-x}{]}_0^1$
$(e^3-1)+(-\frac{1}{e}+1)=\frac{e^4-1}{e}$