Questions: Application of Integrals

Q1 Find the area bounded by the line y equal to 3x, the x-axis and the line x equal to 2, using integration.

Answer (English): The line passes through the origin and stays above the x-axis for positive x, so the region is the triangle with vertices at the origin, at two comma zero and at two comma six. The area is the integral from zero to two of 3x, which is three x squared over two evaluated between the limits, giving six square units. The triangle formula with base two and height six gives the same value, which is a useful check.

Hinglish: Line origin se guzarti hai aur x-axis ke upar hai. Area = ∫_0^2 3x dx = [3x²/2] = 6 square units. Triangle formula se bhi 6 hi aata hai, isliye answer verify ho jaata hai.

Q2 Find the area under the parabola y squared equal to 4x between x equal to 0 and x equal to 1.

Answer (English): On the given interval the upper branch is y equal to two root x and the lower branch is the reflection below the axis. The area between the parabola and the x-axis is the integral of two root x from zero to one, which is four thirds x to the power three halves evaluated between the limits. At one this is four thirds and at zero it is zero, so the area is four thirds square units.

Hinglish: Upper branch y = 2√x hai. Area = ∫_0^1 2√x dx = (4/3)x^{3/2} liya, aur limits lagane par 4/3 square units aata hai.

Q3 Work out the area enclosed between the circle x squared plus y squared equal to 9 and the two vertical lines x equal to minus one and x equal to one.

Answer (English): The upper semicircle is y equal to the square root of nine minus x squared. Because the region is symmetric about the y-axis, the area is twice the integral from zero to one of that square root. The antiderivative is half x root of nine minus x squared plus nine halves inverse sine of x over three. At one this gives half root eight plus nine halves inverse sine of one third, and at zero the first term is zero, so twice the integral gives root eight plus nine inverse sine of one third, which is two root two plus nine inverse sine of one third square units.

Hinglish: Upper semicircle y = √(9−x²) hai aur region symmetric hai, isliye 2∫_0^1 √(9−x²) dx lo. Answer 2√2 + 9 sin^{-1}(1/3) square units aata hai.

Q4 Calculate the area enclosed between the two curves y equal to x squared and y equal to x.

Answer (English): The curves meet where x squared equals x, that is at x equal to zero and x equal to one. On the interval between them the line is above the parabola, which can be checked by testing x equal to one half. The area is the integral of x minus x squared from zero to one, which is x squared over two minus x cubed over three evaluated between the limits, giving one half minus one third, equal to one sixth square units.

Hinglish: Dono curves x = 0 aur x = 1 par milte hain, aur beech mein line upar hai. Area = ∫_0^1 (x − x²) dx = 1/2 − 1/3 = 1/6 square units.

Q5 Find the area between the parabolas y squared equal to 4x and x squared equal to 4y.

Answer (English): Solve the two equations together. Substituting x squared equal to 4y into the first equation gives x to the fourth equal to 64x, whose positive solution is x equal to 4, so the curves meet at the origin and at four comma four. On the interval between zero and four the upper curve is the first parabola written as y equal to two root x, and the lower curve is y equal to x squared over four. The integral of the difference gives sixteen thirds square units after simplification.

Hinglish: Dono equations solve karo; intersection (0,0) aur (4,4) aate hain. Upper curve 2√x aur lower x²/4 hai. Difference integrate karne par 16/3 square units aata hai.

Q6 Using horizontal strips, find the area enclosed by the parabola y squared equal to 4x and its latus rectum.

Answer (English): The latus rectum of y squared equal to four a x is the line x equal to a. Using horizontal strips, y runs from minus two a to two a and the right boundary is the line x equal to a while the left boundary is x equal to y squared over four a. The area is the integral of a minus y squared over four a with respect to y between the limits. The result is eight a squared over three, which is the standard answer for this region.

Hinglish: Latus rectum x = a hota hai. Horizontal strips mein y ke limits −2a se 2a hain, right boundary x = a aur left x = y²/4a. Area nikaalne par 8a²/3 aata hai.

Q7 Find the area of the smaller region cut off from the circle x squared plus y squared equal to 4 by the line x plus y equal to 2.

Answer (English): The line meets the circle at two comma zero and zero comma two, and the smaller region is the part of the first quadrant below the line. Its area is the area under the circle in the first quadrant minus the area of the right triangle with legs two and two. The quarter circle has area pi, the triangle has area two, so the smaller region has area pi minus two square units.

Hinglish: Line circle ko (2,0) aur (0,2) par milti hai. Chhota region first quadrant ke circle area (π) mein se triangle (2) ghata kar milta hai, isliye answer π − 2 square units hai.

Q8 Find the area between the curve y equal to x squared, the x-axis and the lines x equal to minus 1 and x equal to 2.

Answer (English): The parabola is above the x-axis except at the origin, so on the whole interval from minus one to two the integrand is simply x squared. The integral is x cubed over three between the limits, which gives eight thirds at the upper limit and minus one third at the lower limit. The difference is three square units. No split is needed because the curve only touches the axis at the origin and does not cross it.

Hinglish: Parabola x-axis ko sirf origin par touch karta hai, cross nahi karta, isliye poore interval par x² integrate karo. Answer [x³/3] = 8/3 − (−1/3) = 3 square units.

Q9 The area bounded by the line y equal to mx with m positive, the circle x squared plus y squared equal to 4 and the x-axis in the first quadrant is pi by three square units. Find m.

Answer (English): The line y equal to m x makes an angle whose tangent is m with the x-axis, and the region is the sector between the x-axis and the line inside the quarter circle. The area of a sector of a circle is half r squared times the angle, and here r is two, so the area is twice the angle. Setting twice the angle equal to pi by three gives the angle equal to pi by six, so the tangent of the angle, and therefore m, is one over root three.

Hinglish: Region quarter circle ka sector hai. Sector area = (1/2)r²θ = 2θ, aur yeh π/3 diya hai, isliye θ = π/6. Isliye m = tan(π/6) = 1/√3.

Q10 Using integration, work out the area of the triangle with vertices minus one comma two, one comma five and three comma four.

Answer (English): The three sides are the lines joining the vertices. The area can be computed by integrating the upper boundary minus the lower boundary over the x-range from minus one to three, splitting the interval at x equal to one where the upper side changes. The two integrals together give four square units, which agrees with the determinant formula for the area of a triangle with the given vertices.

Hinglish: Teen sides ki equations nikaalo, upper minus lower integrate karo aur x = 1 par split karo. Answer 4 square units aata hai, jo triangle area formula se bhi match karta hai.

The quiz on this page covers areas under curves, areas between curves, horizontal strips and the standard circle and ellipse areas. Attempt it after writing the ten answers above.