Calculus Alıştırma Soruları — JAMB UTME — Sayfa 8

Soru 141Soru →

When differentiating the function $f(x) = x^2 - 3x$ from first principles, what is the simplified expression for the difference quotient $\frac{f(x+h) - f(x)}{h}$ for $h \neq 0$ before taking the limit as $h \to 0$ ?

Cevabı ve açıklamayı göster

Cevap: 2x + h - 3; 2x - 3 + h; h + 2x - 3; 2x+h-3; 2x-3+h

Cevap

2x + h - 3

Expanding

f(x+h) = (x+h)^2 - 3(x+h) = x^2 + 2xh + h^2 - 3x - 3h

and subtracting

f(x) = x^2 - 3x

yields

2xh + h^2 - 3h

. Factoring and dividing by

h

gives the simplified difference quotient

2x + h - 3

.

Adım Adım Çözüm

1

Set up the difference quotient expression

\frac{f(x+h) - f(x)}{h} = \frac{[(x+h)^2 - 3(x+h)] - (x^2 - 3x)}{h}

Substitute

(x+h)

into the function definition to evaluate

f(x+h)

.

2

Expand terms in the numerator

\frac{(x^2 + 2xh + h^2 - 3x - 3h) - x^2 + 3x}{h}

Expand

(x+h)^2 = x^2 + 2xh + h^2

and distribute

-3

across

(x+h)

as well as the negative sign across

f(x)

.

3

Combine like terms in the numerator

\frac{2xh + h^2 - 3h}{h}

The terms

x^2 - x^2 = 0

and

-3x + 3x = 0

cancel out.

4

Divide numerator terms by

h

2x + h - 3

Factor out

h

from the numerator to get

\frac{h(2x + h - 3)}{h} = 2x + h - 3

for

h \neq 0

.

Anahtar Kavram

Difference Quotient in First Principles Differentiation

Tahmini Süre:1m 30s

Soru 142Soru →

What is the area of the region bounded by the curve $y = 6x - x^2$ and the line $y = 2x$ ?

\frac{32}{3}

square units

\frac{80}{3}

square units

\frac{160}{3}

square units

\frac{64}{3}

square units

Cevabı ve açıklamayı göster

Cevap: $\frac{32}{3}$ square units

Cevap

\frac{32}{3}

square units

Equating

y = 6x - x^2

and

y = 2x

gives intersection points

x = 0

and

x = 4

. Integrating the upper curve minus the lower line,

\int_{0}^{4} (4x - x^2) \, dx = \left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4} = 32 - \frac{64}{3} = \frac{32}{3}

square units.

Adım Adım Çözüm

1

Find the points of intersection between the curve and the line

x = 0

and

x = 4

Set

6x - x^2 = 2x \implies 4x - x^2 = 0 \implies x(4 - x) = 0

to find the integration bounds.

2

Set up the definite integral for the area between the two curves

A = \int_{0}^{4} ((6x - x^2) - 2x) \, dx = \int_{0}^{4} (4x - x^2) \, dx

The area between two functions

f(x)

and

g(x)

from

x=a

to

x=b

is given by

\int_{a}^{b} (f(x) - g(x)) \, dx

where

f(x) \ge g(x)

on

[a, b]

.

3

Compute the indefinite integral of the integrand

\int (4x - x^2) \, dx = 2x^2 - \frac{x^3}{3}

Apply the standard power rule of integration

\int x^n \, dx = \frac{x^{n+1}}{n+1}

.

4

Evaluate the definite integral using the limits

0

and

4

\left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4} = \left( 2(4)^2 - \frac{4^3}{3} \right) - 0 = 32 - \frac{64}{3} = \frac{32}{3}

square units

Substitute the upper limit

x = 4

and lower limit

x = 0

and simplify fractions.

Anahtar Kavram

Area Between Two Curves

Tahmini Süre:2m 0s

Soru 143Soru →

A curve has a gradient function defined by $\frac{dy}{dx} = 12x^3 - 6\sin(3x) + 4$ . If the curve passes through the point $(0, 15)$ , what is the value of the constant of integration $C$ ?

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Cevap: 13

Cevap

The constant of integration

C

is 13.

Integrating

12x^3 - 6\sin(3x) + 4

with respect to

x

yields

y = 3x^4 + 2\cos(3x) + 4x + C

. Substituting

x = 0

and

y = 15

gives

15 = 3(0)^4 + 2\cos(0) + 4(0) + C = 2 + C

, which simplifies to

C = 13

.

Adım Adım Çözüm

1

Integrate the gradient function to obtain the general equation of the curve.

y = 3x^4 + 2\cos(3x) + 4x + C

The integral of

12x^3

is

3x^4

, the integral of

-6\sin(3x)

is

+2\cos(3x)

, and the integral of

4

is

4x

.

2

Apply the initial condition

(0, 15)

to solve for

C

.

15 = 3(0)^4 + 2\cos(0) + 4(0) + C

Substitute

x = 0

and

y = 15

into the integrated equation.

3

Calculate the numerical value of

C

.

C = 13

Since

\cos(0) = 1

, the equation becomes

15 = 2 + C

, yielding

C = 13

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions with Initial Conditions

Soru 144Soru →

What is the value of the definite integral $\int_{1}^{3} (4x - 1) \, dx$ ?

14

15

16

12

Cevabı ve açıklamayı göster

Cevap: 14

Cevap

14

The antiderivative of

4x - 1

is

2x^2 - x

. Evaluating at the upper boundary

x = 3

gives

2(3)^2 - 3 = 15

, and at the lower boundary

x = 1

gives

2(1)^2 - 1 = 1

. Subtracting the lower limit evaluation from the upper limit evaluation yields

15 - 1 = 14

.

Adım Adım Çözüm

1

Find the indefinite integral of the integrand

4x - 1

.

\int (4x - 1) \, dx = 2x^2 - x

Using the power rule of integration,

\int x^n \, dx = \frac{x^{n+1}}{n+1}

.

2

Substitute the upper limit

x = 3

into the antiderivative.

2(3)^2 - (3) = 2(9) - 3 = 18 - 3 = 15

Evaluating

F(b)

for

F(x) = 2x^2 - x

at

b = 3

.

3

Substitute the lower limit

x = 1

into the antiderivative.

2(1)^2 - (1) = 2 - 1 = 1

Evaluating

F(a)

for

F(x) = 2x^2 - x

at

a = 1

.

4

Subtract the lower limit result from the upper limit result.

15 - 1 = 14

By the Fundamental Theorem of Calculus,

\int_{a}^{b} f(x) \, dx = F(b) - F(a)

.

Anahtar Kavram

Evaluation of Definite Integrals using the Fundamental Theorem of Calculus

Tahmini Süre:1m 0s

Soru 145Soru →

What is the value of $y$ at the local minimum stationary point of the curve $y = x^3 - 3x^2 - 9x + 15$ for $x > 0$ ?

Cevabı ve açıklamayı göster

Cevap: -12

Cevap

The value of

y

at the local minimum stationary point is

-12

.

To find the local minimum point of

y = x^3 - 3x^2 - 9x + 15

, set the derivative

\frac{dy}{dx} = 3x^2 - 6x - 9

equal to

0

. Solving

3(x - 3)(x + 1) = 0

yields

x = 3

for

x > 0

. The second derivative

\frac{d^2y}{dx^2} = 6x - 6

equals

12 > 0

at

x = 3

, confirming a local minimum. Substituting

x = 3

back into the original curve equation yields

y = (3)^3 - 3(3)^2 - 9(3) + 15 = -12

.

Adım Adım Çözüm

1

Differentiate

y = x^3 - 3x^2 - 9x + 15

with respect to

x

\frac{dy}{dx} = 3x^2 - 6x - 9

Stationary points occur where the slope (first derivative) is zero.

2

Set

\frac{dy}{dx} = 0

and solve for

x

x = 3 \text{ or } x = -1

Factoring

3(x^2 - 2x - 3) = 0

gives

(x - 3)(x + 1) = 0

.

3

Apply the second derivative test at

x = 3

(since

x > 0

)

\frac{d^2y}{dx^2} = 6(3) - 6 = 12 > 0

A positive second derivative confirms that

x = 3

is a local minimum.

4

Evaluate

y

at

x = 3

in the original curve equation

y = (3)^3 - 3(3)^2 - 9(3) + 15 = -12

Substituting

x = 3

into

y(x)

gives the

y

-coordinate of the minimum point.

Anahtar Kavram

Stationary Points and Local Minima of Polynomial Curves

Soru 146Soru →

If $y = \int (4x^3 - 2\sin(x)) \, dx$ and $y = 5$ when $x = 0$ , what is the value of the constant of integration $C$ ?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The constant of integration

C

is 3.

Integrating

4x^3 - 2\sin(x)

gives

y = x^4 + 2\cos(x) + C

. Substituting

x = 0

and

y = 5

yields

5 = 0 + 2(1) + C

, so

C = 3

.

Adım Adım Çözüm

1

Integrate the polynomial and trigonometric terms

y = x^4 + 2\cos(x) + C

The integral of

4x^3

with respect to

x

is

x^4

, and the integral of

-2\sin(x)

is

2\cos(x)

.

2

Apply the given initial condition

y(0) = 5

5 = 0^4 + 2\cos(0) + C \implies 5 = 2 + C

Evaluating the expression at

x = 0

uses the trigonometric property

\cos(0) = 1

.

3

Solve the linear equation for

C

C = 3

Subtracting 2 from both sides isolates

C

.

Anahtar Kavram

Determining the Constant of Integration from Initial Boundary Conditions

Soru 147Soru →

What is the area of the region bounded by the curve $y = \sin x$ and the straight line $y = \frac{2}{\pi}x$ in the first quadrant for $0 \leq x \leq \frac{\pi}{2}$ ?

1 - \frac{\pi}{4}

square units

1 + \frac{\pi}{4}

square units

1 - \frac{\pi}{2}

square units

\frac{\pi}{4}

square units$

Cevabı ve açıklamayı göster

Cevap: $1 - \frac{\pi}{4}$ square units

Cevap

The area of the enclosed region is

1 - \frac{\pi}{4}

square units.

The bounded area is calculated by taking the definite integral of the upper curve minus the lower curve over the interval

[0, \pi/2]

. The upper boundary is

y = \sin x

and the lower boundary is

y = \frac{2}{\pi}x

. Computing

\int_{0}^{\pi/2} (\sin x - \frac{2}{\pi}x) dx

gives

[-\cos x - \frac{x^2}{\pi}]_{0}^{\pi/2} = (0 - \frac{\pi}{4}) - (-1 - 0) = 1 - \frac{\pi}{4}

square units.

Adım Adım Çözüm

1

Set up the definite integral for the area between the two curves

A = \int_{0}^{\pi/2} \left(\sin x - \frac{2}{\pi}x\right) dx

In the interval

[0, \pi/2]

, the curve

y = \sin x

lies above the straight line

y = \frac{2}{\pi}x

.

2

Find the antiderivative of each term

\int \sin x \, dx = -\cos x \quad \text{and} \quad \int \frac{2}{\pi}x \, dx = \frac{x^2}{\pi}

Integration rules for basic trigonometric and power functions.

3

Evaluate the definite integral from lower limit

x = 0

to upper limit

x = \frac{\pi}{2}

A = \left[-\cos x - \frac{x^2}{\pi}\right]_{0}^{\pi/2} = \left(-\cos\frac{\pi}{2} - \frac{(\pi/2)^2}{\pi}\right) - \left(-\cos 0 - \frac{0^2}{\pi}\right)

Apply the Fundamental Theorem of Calculus:

F(b) - F(a)

.

4

Simplify the numeric expression

A = \left(0 - \frac{\pi}{4}\right) - (-1 - 0) = -\frac{\pi}{4} + 1 = 1 - \frac{\pi}{4}

Since

\cos(\pi/2) = 0

and

\cos(0) = 1

.

Anahtar Kavram

Area bounded between curves using definite integration

Tahmini Süre:2m 0s

Soru 148Soru →

What is the indefinite integral $\int (5x^4 - 3\sin(x)) \, dx$ ?

x^5 + 3\cos(x) + C

x^5 - 3\cos(x) + C

x^5 + 3\cos(x)

20x^3 - 3\cos(x) + C

Cevabı ve açıklamayı göster

Cevap: $x^5 + 3\cos(x) + C$

Cevap

x^5 + 3\cos(x) + C

Integrating

5x^4

gives

x^5

, and integrating

-3\sin(x)

gives

+3\cos(x)

since

\int \sin(x) \, dx = -\cos(x)

. Adding the arbitrary constant of integration

C

results in

x^5 + 3\cos(x) + C

.

Adım Adım Çözüm

1

Integrate the polynomial term

5x^4

\int 5x^4 \, dx = \frac{5x^{4+1}}{4+1} = x^5

Apply the power rule for integration:

\int x^n \, dx = \frac{x^{n+1}}{n+1}

.

2

Integrate the trigonometric term

-3\sin(x)

\int -3\sin(x) \, dx = -3(-\cos(x)) = 3\cos(x)

The integral of

\sin(x)

with respect to

x

is

-\cos(x)

.

3

Combine the terms and add the constant of integration

x^5 + 3\cos(x) + C

An indefinite integral requires an arbitrary constant

C

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 149Soru →

If $y = x^3 \sin(2x)$ , what is $\frac{dy}{dx}$ ?

3x^2 \sin(2x) + 2x^3 \cos(2x)

3x^2 \sin(2x) + x^3 \cos(2x)

6x^2 \cos(2x)

3x^2 \sin(2x) - 2x^3 \cos(2x)

Cevabı ve açıklamayı göster

Cevap: $3x^2 \sin(2x) + 2x^3 \cos(2x)$

Cevap

The derivative

\frac{dy}{dx}

is

3x^2 \sin(2x) + 2x^3 \cos(2x)

.

Applying the product rule

\frac{d}{dx}[uv] = u \frac{dv}{dx} + v \frac{du}{dx}

with

u = x^3

and

v = \sin(2x)

yields

x^3(2\cos(2x)) + \sin(2x)(3x^2)

, which simplifies to

3x^2 \sin(2x) + 2x^3 \cos(2x)

.

Adım Adım Çözüm

1

Identify the two component functions for the Product Rule

Let

u = x^3

and

v = \sin(2x)

, so

y = uv

.

The function is a product of an algebraic term and a trigonometric term.

2

Find the derivative of each component function

\frac{du}{dx} = 3x^2 and \frac{dv}{dx} = 2\cos(2x) using the chain rule for \sin(2x).

The derivative of \sin(2x) with respect to 2x is \cos(2x), multiplied by the derivative of 2x which is 2.

3

Apply the Product Rule formula \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

\frac{dy}{dx} = x^3 \cdot 2\cos(2x) + \sin(2x) \cdot 3x^2 = 3x^2 \sin(2x) + 2x^3 \cos(2x).

Combining the differentiated terms according to the product rule gives the final derivative.

Anahtar Kavram

Product Rule combined with Chain Rule for trigonometric functions

Tahmini Süre:1m 30s

Soru 150Soru →

What is the value of the definite integral $\int_{0}^{2} (3x^2 + 2) \, dx$ ?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

12

To evaluate

\int_{0}^{2} (3x^2 + 2) \, dx

, integrate

3x^2 + 2

to get

x^3 + 2x

. Substituting the limits gives

(2^3 + 2(2)) - (0^3 + 2(0)) = (8 + 4) - 0 = 12

.

Adım Adım Çözüm

1

Integrate the function term by term

\int (3x^2 + 2) \, dx = x^3 + 2x

By the power rule of integration, \int 3x^2 \, dx = x^3 and \int 2 \, dx = 2x.

2

Apply the limits of integration from 0 to 2

[x^3 + 2x]_0^2 = (2^3 + 2(2)) - (0^3 + 2(0)) = 12 - 0 = 12

Substitute the upper limit 2 into the antiderivative and subtract the value obtained by substituting the lower limit 0.

Anahtar Kavram

Evaluation of Definite Integrals

Soru 151Soru →

Using differentiation from first principles, which expression represents the derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$ of the function $f(x) = \frac{2}{x}$ , where $x \neq 0$ ?

-\frac{2}{x^2}

\frac{2}{x^2}

-\frac{2}{x}

-\frac{1}{x^2}

Cevabı ve açıklamayı göster

Cevap: $-\frac{2}{x^2}$

Cevap

-\frac{2}{x^2}

The option stating

-\frac{2}{x^2}

is correct because applying the first principles limit formula

\lim_{h \to 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h}

gives

\lim_{h \to 0} \frac{-2h}{h x(x+h)} = \lim_{h \to 0} \frac{-2}{x(x+h)} = -\frac{2}{x^2}

.

Adım Adım Çözüm

1

Set up the difference quotient definition of derivative from first principles.

\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

First principles definition states that the derivative is the limit of the rate of change as

h

approaches zero.

2

Substitute

f(x) = \frac{2}{x}

and

f(x+h) = \frac{2}{x+h}

into the formula.

\frac{f(x+h) - f(x)}{h} = \frac{\frac{2}{x+h} - \frac{2}{x}}{h}

Replacing the function expressions yields the explicit algebraic fraction.

3

Combine the fractions in the numerator over a common denominator.

\frac{2x - 2(x+h)}{x(x+h) \cdot h} = \frac{2x - 2x - 2h}{h x(x+h)} = \frac{-2h}{h x(x+h)}

Simplifying the numerator yields a factor of

h

that can be cancelled with the denominator.

4

Cancel the common factor

h

and evaluate the limit as

h \to 0

.

\lim_{h \to 0} \frac{-2}{x(x+h)} = \frac{-2}{x(x+0)} = -\frac{2}{x^2}

Dividing out

h

allows direct substitution of

h = 0

without producing an indeterminate form.

Anahtar Kavram

Differentiation from First Principles for Reciprocal Functions

Tahmini Süre:1m 30s

Soru 152Soru →

What are the coordinates of the point on the curve $y = 2x^2 - 5x + 1$ where the tangent line is perpendicular to the line $x + 3y - 4 = 0$ ?

(2, -1)

(2, 1)

(1, -2)

(-2, 19)

Cevabı ve açıklamayı göster

Cevap: $(2, -1)$

Cevap

The point on the curve is

(2, -1)

.

Rearranging the line equation

x + 3y - 4 = 0

gives a gradient of

-\frac{1}{3}

. The tangent line is perpendicular, so its gradient must be

3

. Differentiating

y = 2x^2 - 5x + 1

gives

\frac{dy}{dx} = 4x - 5

. Setting

4x - 5 = 3

gives

x = 2

. Substituting

x = 2

into the curve equation yields

y = 2(2)^2 - 5(2) + 1 = -1

. Thus, the point is

(2, -1)

.

Adım Adım Çözüm

1

Determine the gradient of the given line.

Rearranging

x + 3y - 4 = 0

into slope-intercept form gives

y = -\frac{1}{3}x + \frac{4}{3}

, so the gradient is

m_1 = -\frac{1}{3}

.

The slope of a linear equation

Ax + By + C = 0

is

-\frac{A}{B}

.

2

Calculate the gradient of the tangent line.

Since the tangent line is perpendicular to the given line, its gradient is

m_T = -\frac{1}{m_1} = 3

.

Perpendicular lines have gradients whose product is

-1

(

m_1 \cdot m_2 = -1

).

3

Find the derivative of the curve and set it equal to the tangent gradient.

\frac{dy}{dx} = \frac{d}{dx}(2x^2 - 5x + 1) = 4x - 5

. Setting

4x - 5 = 3

yields

4x = 8 \implies x = 2

.

The derivative

\frac{dy}{dx}

gives the gradient of the tangent to the curve at any point

x

.

4

Substitute the

x

-coordinate into the original curve equation to find

y

.

y = 2(2)^2 - 5(2) + 1 = 8 - 10 + 1 = -1

.

The point lies on the curve, so its coordinates must satisfy the curve's equation.

Anahtar Kavram

Tangents and Normals to Curves

Soru 153Soru →

By definition, the derivative of the function $f(x) = \frac{4}{x}$ at $x = 2$ is given by the limit of the difference quotient $\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$ . What is the value of this limit?

-1

1

-2

-4

Cevabı ve açıklamayı göster

Cevap: $-1$

Cevap

-1

Substituting

f(2+h) = \frac{4}{2+h}

and

f(2) = 2

into the difference quotient gives

\frac{\frac{4}{2+h} - 2}{h} = \frac{-2h}{h(2+h)} = \frac{-2}{2+h}

. Evaluating the limit as

h \to 0

yields

\frac{-2}{2} = -1

.

Adım Adım Çözüm

1

Calculate

f(2)

and

f(2+h)

f(2) = \frac{4}{2} = 2

and

f(2+h) = \frac{4}{2+h}

These are the two values needed for the difference quotient numerator.

2

Subtract

f(2)

from

f(2+h)

and find a common denominator

f(2+h) - f(2) = \frac{4}{2+h} - 2 = \frac{4 - 2(2+h)}{2+h} = \frac{4 - 4 - 2h}{2+h} = \frac{-2h}{2+h}

Simplifying the numerator expression algebraically.

3

Divide the numerator by

h

to form the difference quotient

\frac{\frac{-2h}{2+h}}{h} = \frac{-2}{2+h}

Canceling the common factor

h

in the numerator and denominator.

4

Evaluate the limit as

h \to 0

\lim_{h \to 0} \frac{-2}{2+h} = \frac{-2}{2+0} = -1

Direct substitution of

h = 0

into the simplified expression gives the instantaneous rate of change.

Anahtar Kavram

Differentiation of reciprocal functions from first principles

Tahmini Süre:1m 30s

Soru 154Soru →

If $y = e^{3x}\cos(2x) + \ln(x + 1)$ , calculate the value of $\frac{dy}{dx}$ at $x = 0$ .

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The derivative evaluated at

x = 0

is 4.

Differentiating

y = e^{3x}\cos(2x) + \ln(x + 1)

with respect to

x

yields

\frac{dy}{dx} = 3e^{3x}\cos(2x) - 2e^{3x}\sin(2x) + \frac{1}{x+1}

. Evaluating this derivative at

x = 0

gives

3(1)(1) - 2(1)(0) + 1 = 4

.

Adım Adım Çözüm

1

Differentiate the product

u(x) = e^{3x}\cos(2x)

using the product rule and chain rule.

\frac{du}{dx} = 3e^{3x}\cos(2x) - 2e^{3x}\sin(2x)

By the product rule

\frac{d}{dx}(uv) = u'v + uv'

, where

\frac{d}{dx}(e^{3x}) = 3e^{3x}

and

\frac{d}{dx}(\cos(2x)) = -2\sin(2x)

.

2

Differentiate the logarithmic term

v(x) = \ln(x + 1)

.

\frac{dv}{dx} = \frac{1}{x + 1}

The derivative of

\ln(g(x))

is

\frac{g'(x)}{g(x)}

.

3

Combine the terms to write the complete derivative

\frac{dy}{dx}

.

\frac{dy}{dx} = 3e^{3x}\cos(2x) - 2e^{3x}\sin(2x) + \frac{1}{x + 1}

The derivative of a sum of functions is the sum of their individual derivatives.

4

Evaluate

\frac{dy}{dx}

at

x = 0

.

\frac{dy}{dx}\Big|_{x=0} = 3e^0\cos(0) - 2e^0\sin(0) + \frac{1}{0 + 1} = 3(1)(1) - 2(1)(0) + 1 = 4

Substitute

x = 0

using

e^0 = 1

,

\cos(0) = 1

, and

\sin(0) = 0

.

Anahtar Kavram

Differentiation of Trigonometric, Exponential, and Logarithmic Functions

Tahmini Süre:1m 30s

Soru 155Soru →

What is the $y$ -intercept of the normal line to the curve $y = x^3 - 3x^2 + 4x - 1$ at the point where $x = 1$ ?

2

0

-1

1

Cevabı ve açıklamayı göster

Cevap: $2$

Cevap

The

y

-intercept of the normal line is

2

.

Evaluating

y = x^3 - 3x^2 + 4x - 1

at

x = 1

yields

y = 1

. Differentiating gives

\frac{dy}{dx} = 3x^2 - 6x + 4

, which evaluates to

1

at

x = 1

. Since the normal is perpendicular to the tangent, its gradient is

-1

. Substituting into

y - 1 = -1(x - 1)

yields

y = -x + 2

, giving a

y

-intercept of

2

.

Adım Adım Çözüm

1

Find the

y

-coordinate of the point of tangency.

At

x = 1

,

y = (1)^3 - 3(1)^2 + 4(1) - 1 = 1 - 3 + 4 - 1 = 1

. Point of contact is

(1, 1)

.

The point must lie on the curve.

2

Differentiate

y

with respect to

x

to find the gradient function.

\frac{dy}{dx} = 3x^2 - 6x + 4.

The first derivative represents the gradient of the tangent to the curve.

3

Calculate the gradient of the tangent and normal at

x = 1

.

Tangent gradient

m_t = 3(1)^2 - 6(1) + 4 = 1

. Normal gradient

m_n = -\frac{1}{m_t} = -1

.

The normal line is perpendicular to the tangent line, so

m_n \cdot m_t = -1

.

4

Determine the equation of the normal line and find its

y

-intercept.

Using

y - y_1 = m_n(x - x_1) \implies y - 1 = -1(x - 1) \implies y = -x + 2

. Setting

x = 0

gives

y = 2

.

The

y

-intercept occurs where the line crosses the

y

-axis (

x = 0

).

Anahtar Kavram

Tangents and Normals to Curves

Soru 156Soru →

Calculate the area of the finite region bounded by the parabola $y = 3x^2 - 12x + 9$ and the $x$ -axis.

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The area of the bounded region is 4 square units.

Finding the x-intercepts of

y = 3x^2 - 12x + 9

gives

x = 1

and

x = 3

. Integrating

-y

from

1

to

3

gives

\left[-x^3 + 6x^2 - 9x\right]_{1}^{3} = 0 - (-4) = 4

square units.

Adım Adım Çözüm

1

Determine the limits of integration by finding the x-intercepts of the curve.

x = 1

and

x = 3

The bounded region lies between the points where the curve intersects the x-axis (

y = 0

).

2

Set up the definite integral with the correct integrand sign.

A = \int_{1}^{3} (9 - 12x + 3x^2) \, dx = \int_{1}^{3} (-3x^2 + 12x - 9) \, dx

Since

y \le 0

on

[1, 3]

, negating the function ensures the calculated area is positive.

3

Integrate term-by-term and evaluate between upper limit 3 and lower limit 1.

\left[-x^3 + 6x^2 - 9x\right]_{1}^{3} = (0) - (-4) = 4

Applying the Fundamental Theorem of Calculus yields the exact value of 4.

Anahtar Kavram

Area bounded by a curve and the x-axis lying below the x-axis

Soru 157Soru →

Find the equation of the normal to the curve $y = 2\sin x - \cos x$ at the point where $x = 0$ .

x + 2y + 2 = 0

2x - y - 1 = 0

x + 2y - 2 = 0

x - 2y - 2 = 0

Cevabı ve açıklamayı göster

Cevap: $x + 2y + 2 = 0$

Cevap

x + 2y + 2 = 0

At

x = 0

, the

y

-coordinate is

2\sin(0) - \cos(0) = -1

. Evaluating the derivative

\frac{dy}{dx} = 2\cos x + \sin x

at

x = 0

yields a tangent slope of

2

. Since the normal is perpendicular to the tangent, its gradient is

-\frac{1}{2}

. Substituting the point

(0, -1)

and slope

-\frac{1}{2}

into the line formula yields

x + 2y + 2 = 0

.

Adım Adım Çözüm

1

Find the

y

-coordinate of the point of contact

At

x = 0

,

y = 2\sin(0) - \cos(0) = 0 - 1 = -1

. The point is

(0, -1)

.

The line equation requires a point

(x_1, y_1)

on the curve.

2

Differentiate the curve to find

\frac{dy}{dx}

\frac{dy}{dx} = 2\cos x - (-\sin x) = 2\cos x + \sin x

.

The derivative gives the gradient function of the curve.

3

Calculate the gradient of the tangent and normal at

x = 0

Tangent gradient

m_t = 2\cos(0) + \sin(0) = 2(1) + 0 = 2

. Normal gradient

m_n = -\frac{1}{m_t} = -\frac{1}{2}

.

The normal line is perpendicular to the tangent line.

4

Form the equation of the normal line

y - (-1) = -\frac{1}{2}(x - 0) \implies y + 1 = -\frac{1}{2}x \implies 2y + 2 = -x \implies x + 2y + 2 = 0

.

Apply the point-slope form

y - y_1 = m(x - x_1)

.

Anahtar Kavram

Equation of Normal to a Curve

Tahmini Süre:1m 30s

Soru 158Soru →

A curve has a gradient function defined by $\frac{dy}{dx} = 6x^2 + 8\sin(4x) + 3$ . If the curve passes through the point $(0, 10)$ , what is the value of the constant of integration $C$ ?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The constant of integration

C

is

12

.

Integrating

6x^2 + 8\sin(4x) + 3

with respect to

x

gives

y = 2x^3 - 2\cos(4x) + 3x + C

. Substituting

(0, 10)

into the integrated equation gives

10 = 2(0) - 2(1) + 3(0) + C

, which leads directly to

10 = -2 + C

, so

C = 12

.

Adım Adım Çözüm

1

Integrate the gradient function to obtain the general equation of the curve.

y = 2x^3 - 2\cos(4x) + 3x + C

The antiderivative of

\sin(kx)

is

-\frac{1}{k}\cos(kx)

and the antiderivative of

x^n

is

\frac{x^{n+1}}{n+1}

.

2

Apply the given initial condition

(x, y) = (0, 10)

to solve for

C

.

10 = 2(0)^3 - 2\cos(0) + 3(0) + C \implies 10 = -2 + C

Evaluating at

x = 0

requires evaluating

\cos(0) = 1

, which leaves

-2

from the trigonometric term.

3

Solve the linear equation for

C

.

C = 12

Adding

2

to both sides of

10 = -2 + C

gives

C = 12

.

Anahtar Kavram

Indefinite Integration with Boundary Conditions

Soru 159Soru →

A curve $y = F(x)$ has a gradient function given by $\frac{dy}{dx} = 9x^2 - 8x + 6\cos(3x) - 4\sin(2x)$ . Given that $y(0) = 7$ , determine the value of the constant of integration $C$ when the antiderivative is expressed in the standard form $y = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C$ .

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The value of the constant of integration

C

is 5.

Integrating

\frac{dy}{dx} = 9x^2 - 8x + 6\cos(3x) - 4\sin(2x)

yields

y = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C

. Substituting

x = 0

into the expression gives

y(0) = 0 - 0 + 0 + 2(1) + C = 2 + C

. Equating to

y(0) = 7

gives

2 + C = 7

, which solves to

C = 5

.

Adım Adım Çözüm

1

Integrate the gradient function term-by-term with respect to

x

y = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C

The antiderivative of

9x^2

is

3x^3

, of

-8x

is

-4x^2

, of

6\cos(3x)

is

2\sin(3x)

, and of

-4\sin(2x)

is

2\cos(2x)

.

2

Apply the initial boundary condition

y(0) = 7

3(0)^3 - 4(0)^2 + 2\sin(0) + 2\cos(0) + C = 7 \implies 2 + C = 7

At

x = 0

,

\sin(0) = 0

and

\cos(0) = 1

, making the non-zero constant contribution equal to

2(1) = 2

.

3

Solve for the constant of integration

C

C = 5

Subtracting 2 from both sides of

2 + C = 7

yields

C = 5

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions with Boundary Conditions

Soru 160Soru →

The area of the region bounded by the curve $y = 3x^2 - 4x + 3$ , the $x$ -axis, and the vertical lines $x = 0$ and $x = k$ (where $k > 0$ ) is $18$ square units. What is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The value of

k

is

3

.

The area under the non-negative curve

y = 3x^2 - 4x + 3

from

x = 0

to

x = k

is found by calculating

\int_{0}^{k} (3x^2 - 4x + 3) \, dx = k^3 - 2k^2 + 3k

. Setting this expression equal to

18

gives

k^3 - 2k^2 + 3k - 18 = 0

. By the factor theorem, testing

k = 3

yields

3^3 - 2(3)^2 + 3(3) - 18 = 0

. Factoring as

(k - 3)(k^2 + k + 6) = 0

confirms

k = 3

as the only real solution.

Adım Adım Çözüm

1

Set up the definite integral for the area bounded by the curve and the x-axis

\int_{0}^{k} (3x^2 - 4x + 3) \, dx = 18

The curve

y = 3x^2 - 4x + 3

lies entirely above the x-axis for all real

x

because its leading coefficient is positive and its discriminant

(-4)^2 - 4(3)(3) = -20 < 0

.

2

Evaluate the definite integral in terms of

k

\left[ x^3 - 2x^2 + 3x \right]_0^k = (k^3 - 2k^2 + 3k) - 0 = k^3 - 2k^2 + 3k

Applying the fundamental theorem of calculus by integrating term by term.

3

Form and simplify the polynomial equation

k^3 - 2k^2 + 3k - 18 = 0

Equating the definite integral expression to the given area value of 18.

4

Solve for real values of

k > 0

k = 3

Using the factor theorem on

k^3 - 2k^2 + 3k - 18 = 0

,

k = 3

yields zero (

27 - 18 + 9 - 18 = 0

). Factoring out

(k - 3)

gives

(k - 3)(k^2 + k + 6) = 0

, where

k^2 + k + 6 = 0

has complex roots.

Anahtar Kavram

Determining an unknown boundary limit of a definite integral representing area under a curve

Calculus

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