CA Foundation · Quantitative Aptitude · Differential and Integral Calculus
The value of the definite integral of (3x² + 1) with respect to x from x = 0 to x = 2 is:
The definite integral of 3x² + 1 from 0 to 2 equals 10. The antiderivative is x³ + x, which gives 8 + 2 = 10 at the upper limit, and the value at zero is nil, so the area under the curve over that interval is 10.
- A8
- B10Correct
- C14
- D12
Explanation
The antiderivative of 3x² + 1 is x³ + x. At x = 2 it is 8 + 2 = 10, and at x = 0 it is 0, so the value is 10. Choosing 8 comes from dropping the +1 term, which integrates to x and adds 2.
Did you get it right without looking?
One question tells you little. A timed set on Differential and Integral Calculus shows your real accuracy, how long you take and where you lose marks.
More Differential and Integral Calculus questions
- The profit function of a company is P(x) = −x² + 40x − 100 (in ₹), where x is the output in units. The maximum profit is:
- The demand law for a product is p = 49 − x², where p is the price in ₹ and x is the quantity demanded. If the market price is ₹40, the consu…
- The value of the definite integral of x·e^(2x) with respect to x from 0 to 1 is:
- If y = x² eˣ, then the value of dy/dx at x = 1 is:
- If f(x) = 3x² + 5x + 2, find the derivative f'(x).
- Evaluate the definite integral ∫₀² (6x² + 4x) dx.