Calculus (area)

The equation of a curve is x²y + y² = 6x

the equation of the tangent to the curve at the point with coordinates (1, 2)

• 3x – 5y + 8 = 0

• 2x – 5y + 8 = 0

• 2x – 5y - 8 = 0

• 3x – 5y - 8 = 0

Find the gradient of the curve with equation

2x² − 4xy + 3y² = 3, at the point (2, 1).

• 2

• 3

• 4

• 5

The diagram shows the curve y = (x − 2)² and the line y + 2x = 7, which intersect at points A and B. Find the area of the shaded region.

• $10\ \frac{5}{7}$

• $10\ \frac{3}{4\ }$

• $10\ \frac{1}{3\ }$

• $10\ \frac{2}{3}$

 $\int_{ }^{ }\left(x^{3\ }+\frac{1}{x^3}\right)\ dx$  

•  $\frac{x^{4\ }}{4}-\frac{x^{-2}}{2\ }+c$  

•  $\frac{x^{4\ }}{4}+\frac{x^{-2}}{2}+c$  

•  $\frac{x^4}{4}-\frac{x^2}{2\ }+c$  

•  $\frac{-x^4}{4}+\frac{x^{2\ }}{2\ }+c$  

Find the integral for the area under the curve.

• $\int_{-6}^{-2}\ 2\left(x^2\ +\ 6x\ +\ 10\right)\ dx$

• $\int_{-6}^{-2}\ -2\left(x^2\ +\ 6x\ +\ 10\right)\ dx$

• $\int_{-6}^{-2}\ \left(x^2\ +\ 6x\ +\ 10\right)\ dx$

• $\int_{-6}^{-2}\ -4\left(x^2\ +\ 6x\ +\ 10\right)\ dx$

Find the integral that would find the area of the region enclosed by the curves.

• $\int_0^5\left[\left(-\frac{x^2}{2}+4x-3\right)-\left(-x^2+6x-8\right)\right]dx$

• $\int_1^5\left[\left(-x^2+6x-8\right)-\left(-\frac{x^2}{2}+4x-3\right)\right]dx$

• $\int_1^5\left[\left(-\frac{x^2}{2}+4x-3\right)-\left(-x^2+6x-8\right)\right]dx$

• $\int_0^5\left[\left(-x^2+6x-8\right)-\left(-\frac{x^2}{2}+4x-3\right)\right]dx$

Find the area of the region enclosed by the lines and curves.

y = 3x + 4 and y = x² + 4

• 9

• 93/2

• 9/2

• 18

Find the area of the region enclosed by the lines and curves.
y² = x + 8   and  x = y + 64

• 891/2
• 4913/6
• 4937/6
• 955/2

Find the area of the region enclosed by the lines and curves.
y = -4sin(x)   y = sin(2x)    0 ≤ x ≤ Π 

• 8
• 1/2
• 4
• 16

Tags

CCSS.HSF.TF.B.7

_{Evaluate the integral}

• 30/3
• 31/3
• 29/3
• 32/3