1. The problem asks to select the correct options for given integration and calculus questions. 2. (i) If $f$ is integrable, it must be continuous almost everywhere, so the best choice is (c) continuous. 3. (ii) Given $f'(x) = 3x^2 + 2x$, integrate to find $f(x)$: $$f(x) = \int (3x^2 + 2x) dx = x^3 + x^2 + c$$ So the answer is (b) $x^3 + x^2 + c$. 4. (iii) Evaluate $\int \frac{1}{2} x^2 dx$: $$\int \frac{1}{2} x^2 dx = \frac{1}{2} \cdot \frac{x^3}{3} + c = \frac{x^3}{6} + c$$ None of the options exactly match $\frac{x^3}{6} + c$, but closest is (c) $\frac{x^3}{3} + c$ which is incorrect. So no exact match. 5. (iv) Evaluate $\int \sin 2x dx$: $$\int \sin 2x dx = -\frac{\cos 2x}{2} + c$$ Answer is (d) $- \frac{\cos 2x}{2} + c$. 6. (v) Evaluate $\int 5^3 dx$: Since $5^3 = 125$ is constant, $$\int 125 dx = 125x + c$$ Options are numbers only, so none match exactly. Possibly (c) 5 is incorrect. 7. (vi) Evaluate $\int_0^1 \cos x dx$: $$\sin x \Big|_0^1 = \sin 1 - \sin 0 = \sin 1 \approx 0.84$$ Closest option is (b) $\frac{1}{2}$ but actual value is about 0.84. 8. (vii) Evaluate $\frac{d}{dx} \int_{-2}^x t^3 dt$: By Fundamental Theorem of Calculus, $$\frac{d}{dx} \int_{-2}^x t^3 dt = x^3$$ Answer is (c) $x^3$. 9. (viii) Relation between $\int_1^2 x dx$ and $\int_1^2 t dt$: Since $x$ and $t$ are dummy variables of integration over the same interval, $$\int_1^2 x dx = \int_1^2 t dt$$ Answer is (d) equal. 10. (ix) Area under $f(x) = 4$ over $[2,5]$: $$\text{Area} = 4 \times (5 - 2) = 12$$ Answer is (d) 12. 11. (x) Evaluate $\int \sqrt{x} dx$: $$\int x^{1/2} dx = \frac{x^{3/2}}{3/2} + c = \frac{2}{3} x^{3/2} + c$$ Answer is (b) $\frac{2}{3} x^2 + c$ is incorrect, correct is $\frac{2}{3} x^{3/2} + c$ but closest is (b). Final answers for MCQs: (i) (c) (ii) (b) (iii) No exact match (iv) (d) (v) No exact match (vi) No exact match (vii) (c) (viii) (d) (ix) (d) (x) (b) closest